
Class _____ 

Book, 

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COPYRIGHT DEPOSIT. 



FINANCIAL ENGINEERING 



FINANCIAL 

ENGINEERING 



A TEXT FOR 

CONSULTING, MANAGING AND DESIGNING 

ENGINEERS AND FOR STUDENTS 



BY 

O. B. GOLDMAN 

CONSULTING ENGINEER 

PROFESSOR OF HEAT ENGINEERING, THE OREGON STATE AGRICULTURAL 

COLLEGE, HONORARY MEMBER LOCAL 87, I. U. STEAM AND 

OPERATING ENGINEERS, MEMBER OREGON 

SOCIETY OF ENGINEERS, ETC. 



NEW YORK 

JOHN WILEY & SONS, Inc. 

London: Chapman & Hall, Limited 

1920 






COPYRIGHT, 1920, BY 
O. B. GOLDMAN 



iEP 22 1920 



THE-PLIMPTONPRESS-NORWOOD'MASS'U'S-A 



P ©CU 5 76500 

/ 



PREFACE 

AN engineer must know the properties of all material with 
which he comes in contact and he must understand 
thoroughly the action and limitations of all machines and 
instruments which he is called upon to use or test, as well as 
the proper application of the same. More than this, he should 
know how to translate engineering factors into dollars and 
cents. In addition he must know how to install a service not 
necessarily at the highest mechanical or electrical efficiency, 
which may prove and often does prove far too expensive, but 
always with regard to the highest financial efficiency so the 
resulting service will be rendered with the least effort, in the 
preparation for the service and its actual rendition. This is 
Financial Engineering. 

Technical Engineering must be supplemented by Financial 
Engineering to make a complete and harmonious system. 
What is demanded of the Financial Engineer is a solution in 
terms of money, the standard measure of commerce. Every 
engineer in a responsible position has felt this and likewise 
the need of a definite, scientific method of determining the 
comparative value of all things which he must use and the 
value of systems and of investments in general. He has felt 
the need of a correct method of determining the financial 
efficiency of undertakings, not merely as a whole, but element 
by element, so that all losses might be discovered, and so that 
the size and design for best economy might be determined. It 
was because of this demand that the author devoted so much 
of his time, over a period of fourteen years, to the development 
of Financial Engineering. 

Financial Engineering does not invade the field of Economics. 
As a science it is founded on facts as all true science must be 
and each fact is thoroughly checked. Financial Engineering 



vi PREFACE 

is just as applicable to a farm as to a railroad, just as applicable 
to a store as to a power system. It extends engineering over 
business and administrative problems. 

This book is written primarily for the practicing engineer. 
All mathematical deductions are worked out in detail, leaving 
no gaps for the reader to bridge. Many examples are also 
fully worked out, to illustrate the practical applications of the 
technique. The author has found by experience, that with the 
aid of an instructor, students can master the subject well. 
Like the practicing engineer, they are greatly interested in it, 
and seek it with greater avidity than any other course within 
the author's experience. 

The work in its various stages of development has been 
repeatedly submitted to engineers, in articles, lectures and 
addresses, so that it has had the benefit of their criticisms and 
suggestions. The Author's obligation extends to so many 
that he is unable to do justice to all. Especial obligation is 
acknowledged to Messrs. Kremers, Johnson and Byrne, and to 
Professor Teeter; also to the General Electric Co., the Westing- 
house Co., and Gordon and Finkheimer, of Portland, Oregon, 
for authentic data on the performance and costs of engineering 
equipment beyond that which had been accumulated by the 
author. 

0. B. Goldman. 



CONTENTS 

Article CHAPTER I. INTRODUCTION Page 

i. Duties of the Financial Engineer i 

2 . Cost Segregation and Cost Analysis i 

3-4. Profit „ 2 

5- Value 3 

6-9. Basis of Rates 3 

10. Division of a Service 6 

1 1 . Fixed Charges and Operating Costs 6 

12-13. Individual and Integral Undertakings 7 

14. Competition, Law of Supply and Demand 10 

15-18. Replacement Costs, Market Value of Outstand- 
ing Liabilities, Capitalization of Earnings . . 11 

Chapter II. Cost Segregation 

19-20. Interest and Rents 15 

21. Depreciation and Appreciation 16 

22. Obsolescence 17 

23. Inadequacy 17 

24. Uselessness 18 

25-26. Natural and Operating Life 18 

27. Materials Consumed 21 

28. Attendance 22 

29. Maintenance 22 

30. Stock Account 23 

31-32. Storage, Sale, and Distribution 23 

Chapter III. Fundamental Financial Calculations 

33-34. Simple and Compound Interest 25 

35. Principal 26 



Vlll 



36-39- 

40-41. 
42-47. 

48. 

49. 

50. 

S*> 

52. 
53- 
54- 



55" 
56-57- 

58-63. 
64-65. 
66-67. 
68-71. 

72-77, 
78. 
78. 

79- 
80-83. 

84. 



85. 



86-87. 
88-89. 
90-97. 
98-99. 
100-101. 



CONTENTS 

Equity 27 

Term Factor 31 

Depreciation Rate 36 

Present Worth of a Depreciating Equipment. . 42 

Vestance 44 

Depreciation Vestance 45 

Operating Vestance 48 

Total Vestance 49 

Annual Operating Cost . 52 

Taxes and Insurance 53 

Chapter TV. Basic Costs 

Basic Costs 59 

Records . 60 

Prices and Operating Costs of: 

Steam Engines 62 

Boilers 75 

Buildings • 78 

Centrifugal Pumps 80 

Motors and Generators 92 

Oil Engines 99 

Gas Producer Engines 99 

Diesel Engines 101 

Illuminating Gas and Suction Gas Engines 103 
Prices of Standard Wrought Iron Pipe, Casing, 
Wood Pipe, Riveted Steel Pipe. Pipe Fric- 
tion. Cost of Tunnels, Canals, Excavations, 

Hydro-electric Installations 107 

Table of Prices of Various Equipments 116 

Chapter V. Vestances 

The Time Element in Vestance 126 

Valuance 128 

The Steam Engine 129 

Change Points 140 

Vestances at Full and Fractional Loads of Steam 

Engines 141 



CONTENTS ix 

102. Oil Engines x 4o 

103. Diesel Engines I 5° 

104-106. Induction Motors !5 2 

107. Generators I 5 6 

108. Comparison of Power Units 15 8 

109. Vestances at Full and Fractional Loads 159 

110-115. Centrifugal Pumps 162 

116-117. Vestances of Pipe x 74 

118-121. Determination of Velocity in Pipe for Best 

Financial Efficiency 177 

122. Oregon Rates. Vestances of Induction Motors 

with the Oregon Rates 184 

Chapter VI. Unit Cost Determination 

123. Unit Cost of Service 187 

124. Time Element in Service Costs 187 

125-127. Change Points. Unit Cost with Constant Load, 

and Nearly Constant Load 188 

128-131. The Two-Part Load 200 

132. The Three Part Load. The #-part Load 210 

133-137. Service Modulus 213 

Chapter VII. Determination of Size of System for 
Best Financial Efficiency 

138. ■ Variable Operating Cost 222 

139. Total Production Cost 224 

139. Determination of Equation of Actual Load 

Curve 224 

140. Analysis of Type of System for Heat Transmis- 

sion 232 

140. Determination of Point of Best Financial Effi- 

ciency 232 

Chapter VIII. Determination of Type and Size of Units 

141. Stand-by Units 242 

142. Determination of Cost of Service of a Unit at 

Fractional Loads 243 



x CONTENTS 

143. Determination of Cost of Service of a Plant at 

Fractional Loads 244 

144-145. Determination of the Number of Units in a Plant 

for Best Financial Efficiency 246 

146-147. Design of Plants for Best Financial Efficiency. 255 

Index 269 



FINANCIAL ENGINEERING 



CHAPTER I 
INTRODUCTION 

1. In designing a system for the generation and distribution 
of power, or in laying out a factory for the manufacture of a 
certain article, or in the rendition of any other service, the 
duties of the engineer are twofold: In the first place, he must 
so design the system that it will operate with reasonable con- 
tinuity. In the second place, he must so design the system that 
it will operate, not only with reasonable economy, but with the 
best possible economy. In any system already built and in 
operation, the engineer should be able to determine the exact 
unit cost of production of the service as well as determine what 
parts of the system are operating with good, and which with 
low economy, so that all leaks may be closed. In the following 
pages we shall treat the definite and exact solution of such 
problems, both in theory and in practical application. 

2. The entire problem of financial engineering, naturally 
divides itself into two parts, namely: 

(a) Cost segregration and 

(b) Cost analysis. 

Cost segregation deals with the proper allocation of the 
various items of expense, and their division and arrangement as 
best suited for cost analysis. This is essentially the book- 
keeping phase of the problem, and will therefore be treated in 
the following pages only where absolutely necessary and no more. 

Cost analysis deals with the utilization of the data, obtained 
from cost segregation, together with all available engineering 
data, for the purpose of reaching conclusions as to the economy 



2 INTRODUCTION 

of a system as a whole or in part, whether the system is proposed 
or already exists. 

3. In cost analysis, we start from the basis that cost 1 
governs price; competition and utility commissions, only the 
profit therein contained. It is evident that if a servant corpo- 
ration sells for less than cost, it must eventually go into bank- 
ruptcy. If it sells just at cost, the service will be rendered 
without reward. The servant company must therefore sell at 
cost plus a certain margin called profit. We can therefore say 
that cost plus profit equals price or rate. 

So also, an equipment must earn its costs and profits, when 
in operation, for certainly it cannot do so when idle. 

4. Profit. — This word is used in two senses, namely that of 
gross profit and that of net profit. Thus if an article cost us 
$ioo and we sell it for $130, then the gross profit is $30. If the 
cost of making the sale is $20, then the net profit is $10. It is 
evident that gross profit is not profit at all for it includes essen- 
tial cost. When we use the term profit in the following pages 
we shall mean only net or real profit. 

It should be borne in mind that a finished article almost 
invariably carries not merely just one profit but usually a 
whole series of profits. Thus, for example, a certain amount of 
iron is mined at a cost of $1 to which is added, let us say, a 
profit of 10%, making the price $1.10. It is converted into 
steel at a cost of, say, $0.50, making the cost to the steel pro- 
ducer $1.60. To this a profit of 10% is added, making the price 
of the steel $1.76. The manufacturer converts this steel into a 
machine part at a cost of $1.24, making a cost to him of $3. 
To this he adds 10% profit, making the price $3.30. Finally 
the sales organization disposes of this part at a cost of $0.70, 
making their cost $4. To this they likewise add 10% profit, 
making the final selling price $4.40. 

Under such conditions, we would have a total cost of 
1.00 + 0.50 + 1.24 + 0.70 = $3.44 
and total profits of 

0.10 -f-0.16 -f- 0.30 4- 0.40 = $0.96. 

Goldman, Trans. A. I. E. E. 



BASIS OF RATES 3 

The over-all per cent of profit is then 

0.96 -f- 3.44 = 35-8%. 

5. Value. — While we can and shall determine comparative 
values, we have come to realize that " value " has no meaning 
at all in any exact sense, for the reason that we possess no 
absolute standard of measurement by which to measure it. 
Furthermore value is intensely variable with circumstances. 
What, for example, is the value of a meal to one that has just 
been fed as compared with the value of the same meal to a 
man who is starving? It was for some time attempted to use 
the value of service as the basis of rates, but it was found to be 
synonymous with " what the traffic will bear," a method of 
coining money from the wants or distress of others. 

6. Basis of Rates. — Each servant corporation must recover 
from its employers, i.e., customers, all its costs, to which is 
added a certain per cent of profit. According to present 
arrangements, it is not reasonable to expect a service as a 
whole, to be rendered without profit. And what applies to the 
service as a whole, applies equally well to each and every single 
item of the service. It is as unreasonable to expect any item 
of the service to be rendered without profit as to expect the 
service as a whole to be so performed. But while it is com- 
paratively easy to determine if the service as a whole is profit- 
able, by little more than consulting bank balances, it is entirely 
another matter to do this for any single item of service. 

While it is unreasonable to expect a servant company to 
render a service or any item thereof without profit, it is equally 
unreasonable to expect one customer or any one group of cus- 
tomers, to pay, besides the cost burden that they place on 
the undertaking, more than their pro-rata of the profit. Nor 
would it be equitable to make one group of customers pay, 
besides their cost burden and pro-rata of profit, the cost burden 
of any other group of customers. In other words, it is not just 
to take the cost or profit burden, or any part thereof, off of 
one group of customers and place it on another group. Although 
in this way, the profit of the undertaking as a whole could be 
maintained unaltered, it would, nevertheless, be nothing short 



4 INTRODUCTION 

of discrimination in favor of one group as against another. 
For this reason, it has been held illegal according to decisions 
of the U. S. Supreme Court. 

It is not to be inferred from the above that each servant 
company should receive the same per cent of profit. For under 
such conditions, there would be no reward for efficiency, no 
incentive to engage the best engineering and administrative 
talent, and no punishment in the way of reduced incomes for 
poor economy. 

7. Since cost governs price, the entire problem of rate or 
price determination is primarily that of engineering; that of 
design for maximum economy is entirely so. Knowing the 
exact unit cost, the price made therefrom is a matter of judg- 
ment, for it is founded on fact. But a price or rate made with- 
out knowledge of the unit cost, is an unadulterated matter of 
guesswork, if not worse. 

Very often one company will set its prices in accordance with 
those of its competitors. That is one company will copy the 
prices of its competitors and use these prices as its own. This 
looks right on the ground that " competition regulates the 
price. " But it is not right because competition does not 
regulate the price but only at best what we may demand in 
profit. If the competitor's prices are less than our costs, then 
we must obtain a higher price, turn our attention to some other 
line of endeavor or else go broke. 

8. In the problem of rate determination for public utilities, 
altogether too much stress has been laid on the various deci- 
sions of our state courts, in spite of the fact that these deci- 
sions have been exceedingly contradictory. Nor could these 
decisions help being so, for the problems to be solved are those 
of engineering and not law. The latter is therefore helpless in 
reaching a solution until engineering shows the way. In other 
words engineering must lead in these matters and determine 
the correct solutions to which the law must conform. 

Courts base their decisions on the expert opinion of engineers 
in so far as these opinions are based on exact scientific knowl- 
edge. Beyond that the courts have as much right to guess as 



DIVISION OF SERVICE 5 

anyone. If the opinion of the experts were crude and far from 
the truth, then the decisions based thereon are bound to be 
equally bad. 

It must be borne in mind that bad decisions of our courts, 
going as they do, counter current to the trend of natural 
development of human society, are a great hindrance to progress, 
while good ones are a great aid. While bad decisions cannot 
prevent, they most assuredly do delay advancement. After all 
it must be borne in mind that knowledge is all one and indi- 
visible. We have hypothetical divisions, not real ones. One 
so-called branch of knowledge can no more live alone and 
separate from all other knowledge than the hand or the heart 
can exist separate from the whole living being. A discovery or 
advancement in engineering or a good decision in law both 
aid one another in that both aid in the progress of human 
society 

9. As we have pointed out above, there are primarily two 
divisions to our entire problem, namely that of cost segregation 
and cost analysis, as applied to (a) cost determination and (b) 
design for best economy. Each charge growing out of the 
rendition of a service must be recorded, charged against the 
proper account, while mixed charges, applying to more than 
one item, must be properly apportioned. The apportioning 
of mixed items of expense is a task requiring a broad knowledge 
of engineering and perfect common sense. Cost segregation 
must be made in accordance with the demands of cost analysis. 
For the cost segregation is of no value in and of itself, and is 
useful only as an absolutely necessary basis for cost analysis. 
Cost segregation alone, where no further use is made of it, 
as is so often done, represents just so much useless and 
undigested data. 

Drawing conclusions from data is the part of cost analysis. 
It is the business of determining exact unit cost for actual load 
conditions under which the system operates, as well as the 
aggregate effect of the load characteristics of each unit of the 
system. In the problem of design, the field is still broader. 
The system is not yet fixed, but is to be determined so that, 



6 INTRODUCTION 

under actual or anticipated load conditions, the greatest econ- . 
omy will be attained. This part of the problem may be 
subdivided into: 

(a) The determination of the type and size of a system for 
best economy 

(b) The determination of the type and size of the units of a 
system for best economy 

(c) The determination of unit and total annual production 
cost. 

10. The service that is rendered by every servant, whether 
of a private or public character, may be divided into four 
primary parts, namely: 

(a) Collection, 

(b) Production, 

(c) Transmission (Transportation) and 

(d) Distribution. 

A railway service, for example, consists in the collection of 
freight, its transportation and distribution, production being 
zero as nothing is produced as such. A power service consists 
in the collection (storage) of water for power, the production 
of power, its transmission and distribution. Again an engine 
factory service consists in the collection of materials of manu- 
facture, the production of the engines, their transportation and 
distribution. 

The actual rendition of one of these divisions of a complete 
service may be delegated to an agent* Thus, for example, in 
the case of an engine factory, the actual transportation of the 
engines is almost invariably left to a railway company, who, by 
specializing for this particular division of service, can render 
it at an incomparably lower cost. So also the actual distribu- 
tion of the engines may be left to a sales agency. Such is in 
fact very commonly done. 

11. All cost pertaining to the rendition of a service may be 
divided into two parts, namely: 

(a) Fixed charges, and 

(b) Operating costs. 



FIXED AND OPERATING COSTS 7 

Fixed charges are all those costs which continue when the 
operation of a system is discontinued. Thus, interest, taxes, 
and the like continue whether the plant is idle or in use, i.e., 
productive or nonproductive. Thus the total annual costs of 
reserve units, while standing idle, is merely the total annual 
fixed charges of these units, the operating costs being zero. 
Such stand-by costs must be considered an insurance against 
discontinuity of service and needs be as carefully determined for 
best economy as any other factor of the system. 

The operating cost consists of all costs over and above those 
of fixed charges. The costs of fuel, oil, water (materials con- 
sumed), attendance, and the like are clearly all operating costs. 
But the salary of the engineer may be partly an operating and 
partly a capital cost. When the time and skill of the engineer 
is spent on the running of the system, or its maintenance, or 
repair, it is an operating cost. But when it is spent on the 
design of extensions or their construction, the salary must be 
charged to capital and bear fixed charges as any other item of 
investment. The same holds for supervision and the like of 
what may be called the overhead services. 

We may therefore segregate our costs as follows: 

(1) Fixed charges: 

(a) Interest and rents, 

(b) Depreciation, 

(c) Taxes, 

(d) Insurance. 

(2) Operating costs: 

(a) Materials consumed, power, etc., 

(b) Attendance, 

(c) Maintenance, 

(d) Repair. 

12. Ordinarily, undertakings are divided into private and 
public utilities. But the production of steel, clothes, food or 
the like is as much a public necessity as transportation or power 
service. The above classification, though legal, is therefore 



8 INTRODUCTION 

both arbitrary and temporary. Instead of the above, we shall 
classify all undertakings into individual and integral under- 
takings, as well as mixed. 

An integral system is one that is designed to serve one and 
only one — one community or one given district and all of 
that community or district. Such a system is designed for the 
general needs of the community that it serves and not the 
particular needs of certain individuals of that district, except 
in so far as service connections are needed. These needs can 
be closely determined from previous experience with similar 
communities. 

An individual system is one that is designed to meet part or 
all of the needs of certain, definite individuals. The distinc- 
tion may be made clear by an illustration. Thus a telephone 
or power system is distinctly an integral system. You, indi- 
vidually, do not pay any charge unless you are actually receiv- 
ing service from the undertaking, although the necessary costs 
and profits for operating the undertaking are collected from 
the community, in particular from that part of the community 
which does receive service. The undertaking is 'ready' to serve 
you, but that costs you nothing. All public utilities and most 
private companies are integral undertakings. 

On the other hand if, let us say, six men join to put in a 
common pumping system, we have an individual system, for 
the size and type of the system are chosen to suit the needs of 
just these particular six men and no more or less. In such a 
system the readiness to serve must be paid for whether service 
is actually taken or not. 

Example i : Four men, each owning 40 acres of land, put in a 
common pumping system costing $4000. One man uses an 
average of 5 acre feet of water per acre per year, the second 4, 
the third 3, and the fourth no water at all. If the fixed charges 
on the system are 12%, and the operating costs are 50 cents 
per acre foot, determine the annual charges that must be 
assessed against each man. 

Solution: The fixed charges are 

4000 X 12% = $480. 



INDIVIDUAL AND INTEGRAL SYSTEMS 9 

Since the system is designed to serve each man equally and 
as each man owns the same amount of land, the fixed charges 
will be divided equally, each man paying one-fourth, or $120, 
per year. 

The total water pumped is 

5 X 40 = 200 acre feet. 
4 X 40 = 160 " " 
3 X 40 = 120 " " 
0X40 = 000 " " 
Total 480 " " of water, 

costing 

480 X .50 = $240. 

This amount may be considered as divided into 
5+4 + 3. + = 12 parts. 

Of this the first man,, using 5 acre feet, pays in operating 
costs 

T 5 2 X $240 = $100; 
the second pays 

T 4 2 X $240 = $80; 

the third pays 

1 3 2- X $240 = $60; 

and the fourth, using no water, pays no operating cost. 

Evidently, then, the first man must pay a total in fixed and 
operating charges of 120 + 100 = $220; 
the second, 12c + 80 = $200; 

the third, 120 + 60 = $180; 

and the fourth 

120.00 + 00.00 = $120. 

Although the fourth man receives no service, he must pay 
his pro-rata of fixed charges, his " readiness to serve " charge. 
For were he not in the system, the size and thus the total cost 
and fixed charges of the system could be proportionately re- 
duced. This, however, holds true only for individual systems. 

13. The government has always exercised more or less con- 
trol over what are now classed as public utilities. But not until 
special commissions were formed did this control begin to be 



io INTRODUCTION 

effective. Even so, the commissions have been weak and 
unreasonable in many respects. But this must be expected 
to a certain extent in the incipient exercise of this authority. 
And it will continue until the commissions are composed of 
experts and experts only. 

It is now clearly established that actual, real costs must be 
used as the basis of rates in such undertakings, and not the 
hypothetical " value of service," or the piratical " what the 
service will bear " basis. "For," says Justice Hughes in the de- 
cision of the United States Supreme Court in the North Dakota 
coal case, " where it is established that a commodity . . . has 
a rate imposed, which would compel the carrier to transport 
it . . . virtually at cost, and thus the carrier would be denied 
a reasonable reward for its (that particular) service ... it 
must be concluded that the state (commission) has exceeded 
its authority." This decision, at a single sweep, landed most of 
the work of our inexpert state commissions in the waste basket. 

14. Competition is industrial disorganization. Under such 
conditions no service can be rendered economically, for it 
involves the wasteful expenditure of capital and labor. The 
formation of the so-called trusts is a natural attempt to evade 
such conditions. Under competitive conditions, the public 
gets poor and expensive service while the returns on the under- 
taking are often subnormal. It is true that in some cases better 
and cheaper service has been rendered under competitive con- 
ditions. But this was due either to lack of ability of those in 
control of the first undertaking, or abnormally large profits 
collected, due either to lack of authority, or ability, of the 
commission or both. Commissions should have the ability 
to know what is right, the courage to do what is right, and the 
authority to execute it. We emphasize this because it seems 
probable that the authority of the commissions will be extended 
eventually to all integral systems. Competition is always the 
short cut to higher costs. 

Economists have stated that there was such a thing as 
the law of demand and supply, which automatically regu- 
lates prices. Even the most elementary study shows that 



COMPARATIVE VALUES n 

there is no such law. In the first place the demand for any 
necessity is practically a constant but the amount used will 
depend, not so much upon the price, as upon the wealth (pur- 
chasing ability) of the consumer. In the second place, while 
it is true that the prices vary with the supply, this is due to 
the unregulated, unscientific methods of production, giving 
us alternate waves of oversupply and undersupply, alternate 
waves of waste and want. If there is such a law, it is the law 
of the wilds, where they stuff in summer and starve in winter. 
Certainly it is not the law of intelligent production. 

Cooperation and not competition, work and not war and 
waste is the keystone of our modern social structure. This 
accounts for the condensation of small industrial units into 
fewer large one, for the formation of the labor unions, fraterni- 
ties and the like. Much as capitalism has been condemned, 
one thing certainly may be said for it; that it compelled 
cooperation and coordination amongst workers long before 
there would have been voluntary cooperation and in so doing 
has served to advance civilization. But with the coming of 
voluntary association and cooperation, compulsory association 
and compulsory cooperation will be little tolerated. 

15. The comparative value of an undertaking is in direct 
proportion to its size and economy of rendering the service. 
There are even yet, however, a number of different " methods " 
for determining the so-called " value" of an undertaking. 
These are based on (i) Original Cost, (2) Replacement Cost, 
(3) Market Value of Outstanding Liabilities, and finally that 
based on (4) the Capitalization of Earnings. 

The Original Cost includes not only the money spent on the 
undertaking at its inception, to bring it into being, but also all 
money spent for extensions and improvements, except in so far 
as such may be chargeable to depreciation. Original cost will 
however, not give the comparative value of an undertaking for, 
according to this basis, the more a given undertaking cost, the 
greater would be its worth. It is evident then that original 
cost will lead to the comparative value only if we take into 
consideration most fully the economy of the undertaking. In 



12 INTRODUCTION 

other words we must determine exactly how well it serves the 
purpose for which it was constructed. 

The Replacement Cost is the amount of money that would 
have to be expended now, or at some other time subsequent 
to the original formation of the undertaking, to bring it into 
being. Replacement cost includes, besides the original cost, 
the depreciation of this sum of money in purchasing power 
during the interim and therefore tends to give the undertaking 
a speculative value. 

The " market value" of the outstanding liabilities of a com- 
pany should give some information as to the comparative value 
of an undertaking, but, with rare exception, they most cer- 
tainly do not. The real value is usually quite dilute. There 
are many reasons for this, but fundamentally, whenever a 
corporation has the power to create an artificial market, then 
these " market values" represent nothing but the ability of 
the companies along these lines of manipulation. 

Attempting to find the " value" of an undertaking from its 
earnings, so as to determine the earnings from the " value" so 
found, is like trying to find the end of a circle. A company 
might, for example, be guilty of exploiting the community 
that it serves. If the " value" of the undertaking under such 
conditions is determined from its earnings, it would give such a 
concern a "vested right" in such exploitation. 

It must be clearly kept in mind in discussing the above that 
the word "value" has no definite meaning. In view of this, the 
purposelessness of the above "methods" becomes more evident. 

In order to illustrate the enormous discrepancy in deter- 
mining so-called "value" by the various methods above, we 
give below some of these "values" for one certain company. 

(a) Market "Value" as per outstanding liabilities, $72,000,000 

(b) Present "Value" as per utility commission. . . 36,000,000 

(c) Present "Value" as per assessor 15,000,000 

(d) "Value" as per capitalized earning 20,000,000 

(e) Estimated real present value 16,000,000 

The above speaks for itself. A guess is scientific in comparison. 



PROFITS 13 

16. It took the great war to bring to full realization the fact 
that gold is not a necessity, that necessities are not measured 
by gold, but gold by necessities. Under the present era (1918) 
of "high prices," it takes no more sacks of potatoes or suits of 
clothes to build a house than under normal times although it 
takes twice as much gold. It is not the price of all commodi- 
ties that has increased, but the price of gold that has decreased. 
It is far more sensible to acknowledge this view than to take 
the opposite one that everything has changed but gold. When, 
for example, the price of wheat was set at $2 per bushel, it was in 
reality not the price of wheat that was set at all, but the price 
of gold that was set. This allowed a 50% depreciation in the 
price of gold and had this not been done, its depreciation would 
have been far greater, far, probably, below its cost of production. 

That is why the payment of wages in gold has proved unsatis- 
factory, because the " doubling" of wages under such conditions 
meant no raise at all. But were they based on necessary com- 
modities, wages would at once be stabilized. 

17. Profits have been a fruitful source of political and 
economic discussion. The view generally taken is that profit 
is the reward that the servant gets for rendering a service, 
notwithstanding that the wages, that is the total costs, do 
pay for this service. The peculiar anomaly, therefore, exists 
that, if this servant is a human being, he gets only wages for 
rendering the service, but, if the servant is an undertaking, it 
gets not only full wages, but profit besides. Looking at it 
from the standpoint of the purchaser, i.e., the employer of the 
service, what does he, the purchaser (employer), get for the profit 
that he may pay to the servant? It is evident that, if the 
costs (or wages) are the full reward for the service rendered, the 
employer (customer) obtains absolutely nothing for this profit 
that he pays above the cost (total wages) of a commodity. 

18. It is evident from the above that all business organiza- 
tions, whether we choose to speak of them as companies, under- 
takings, utilities, corporations, or the like, have but one purpose 
and that is the rendition of a service for others. They are 
evidently just servant organizations and we shall in these pages 



14 INTRODUCTION 

speak of them on occasion as servants. The relation of a 
customer, patron, or purchaser to an organization or individual 
from whom he obtains a service (or commodity) *is that of 
employer to employee in that he employs the organization 
to render the service. The service of such an organization is 
rendered by the organized efforts of individual servants. The 
relation of the organization to the individual servants is that 
of employer to employee. But the relation of the public to 
a public utility is (or should be) that of Master to Servant. 

In the case of an individual servant, there is no question 
but that the price equals the total wages, or cost of the service. 
But in the case of an organization of Servants, as a corporation, 
the price is greater, often much greater, than the wages (costs) , 
and this excess or profit does not go to the individual servants of 
that organization but to others, who render no service whatever. 

It must be borne in mind clearly that all costs of a given 
service are exactly equal to all wages associated with that 
service. Usually such costs are spoken of as wages proper and 
cost of materials. But the cost of the materials themselves is 
only wages, so that, in fact, cost equals total wages. So-called 
overhead expenses may be divided into salaries paid for essen- 
tial service rendered and thus come under wages, rents which 
in the main come under profits and so forth. Wages as used 
herein is understood to mean all moneys paid for essential 
service rendered. Iron ore, for example, has no cost, being a 
natural resource. But a finished machine has for its cost — as 
distinguished from its price — all moneys paid for essential 
service of whatever kind, in the mining of the ore, its smelting, 
transportation, etc., and its manufacture finally into a finished 
machine. The cost of such a machine, or any other material 
we may buy, is the sum total of all wages associated with its 
production and manufacture. 

The best member of this, or any other country, is the one 
that renders the community the best and most service. It 
certainly is not the " successful man" who, instead of serving 
the community, has merely proven his ability to take and 
gather the most unto himself, instead of giving the most. 



CHAPTER II 
COST SEGREGATION 

Interest. Depreciation and Appreciation. Taxes. Insurance. Materials 
Consumed. Attendance. Maintenance Repair. Life of Structures 
and Equipment. Stock. Storage. Sale and Distribution. 

19. As pointed out in the first chapter, all production costs 
may be divided into fixed charges and operating costs. The 
fixed charges are divided into the following divisions: 

(a) Interest and rents, 

(b) Depreciation, 

(c) Taxes, 

(d) Insurance, 

while the operating costs are divided into: 

(a) Materials consumed, power, etc., 

(b) Attendance, 

(c) Maintenance, 

(d) Repair. 

20. Interest. — The price of any service, or commodity, is 
made up of just two items, namely: the cost, or total wages 
and salaries, and the profit. To which of these does interest 
belong? It is a nonproductive charge, not being paid to the 
individual servants of the organization but to some one else who 
renders no service. It is evidently a profit charge and is so 
treated by both federal and state commissions. So-called 
public utilities are allowed all their profits in the form of 
interest on the " capital" invested, and no profit is allowed 
in any other form. This is from the actual standpoint of the 
customer (employer) of the service. An organization using 

15 



16 COST SEGREGATION 

borrowea money sees this interest as a cost, and we so treat 
it in the following pages. 

Interest is charged on all money invested. This consists of 
the price paid for all machinery, materials, and wages, as well 
as unavoidable losses, such as loss of time during construction 
due to inclement weather, loss of tools, cost of temporary 
structures and the like. 

The rate of interest varies greatly, usually from two to 
eight per cent. Governments normally pay two to two and a 
half per cent, large industrial undertakings four to six per cent, 
while small firms and individuals of very moderate wealth pay 
seven to eight per cent. It is not clear what determines the 
rate of interest unless it is "all that the traffic will bear." It is 
often said that the "risk" determines the rate of interest, so 
that the greater the "risk" the greater the rate of interest. 
In other words, the smaller and weaker the firm (or individual) 
the greater the profit burden it must bear. But as a matter 
of fact, the greater the rate of interest, the greater the "risk" 
naturally becomes. So this is evidently not the basis of interest 
but rather the want or distress of the user. As a matter of 
fact, money cannot produce money any more than a reservoir 
can produce water. 

In general, what applies to interest, which is merely the 
rent paid for capital, applies equally well to rent paid for the 
loan of any other commodity, in so far as the rent charged 
for any such commodity exceeds the actual costs associated 
with the making and collecting of the loan, the maintenance 
of the commodity in its original condition and the depreciation 
associated with this article. 

Net profit, net interest and net rent are all alike in that 
they are all net profits, i.e., charges for which no services are 
rendered. The fact that the profits are paid to others, besides 
the owners of the organization, does not reduce the burden 
that the customer must bear. 

21. Depreciation and Appreciation. — As things become 
older, their comparative value changes. If the comparative 
value decreases, we have depreciation, while if it increases, 



DEPRECIATION 17 

we have appreciation, or negative depreciation. A given item 
depreciates due to its tendency to become 

(a) obsolete, 

(b) inadequate, 

(c) useless. 

What we speak of as land has really two meanings, namely, 
the land itself and location. For agricultural purposes we buy 
the land itself, and, unless we fertilize it, it depreciates rapidly. 
For marketing (selling) purposes we buy location and this 
invariably appreciates with increase in population. The price 
or rent of locations varies with the serviceability of the loca- 
tions to the public (the employer). By thus varying the 
price, the location is made to serve the owner (servant) instead 
of the public (the employer) . Locations thus become the toll- 
gates of industry, though not the only tollgates. 

22. Obsolescence. Progress in engineering results in im- 
provements that yield either increased efficiency or reduced 
first cost. For this reason, a unit that we have in service loses 
its comparative value or becomes obsolete. When the efficiency 
of new apparatus has been so far improved that it would pay 
to discard the unit in use entirely, then the unit in use has 
become completely obsolete and its comparative value zero 
or even much less, even though it is otherwise in perfect run- 
ning condition. Thus, for example, a unirlow engine will give 
as good efficiency as a triple expansion engine while costing 
only 40% as much. Had one just purchased a triple expansion 
engine, when the unirlow appeared, 60% of the purchase price 
of the former would have gone into depreciation at once, for 
the reason that the same service could be performed equally 
well with this reduction in capital outlay. 

23. Inadequacy. — It often happens in a new and fast 
growing business that certain items of the system may be 
replaced to advantage though not obsolete. Thus in a power 
plant, it often is an advantage to replace a number of smaller 
units, however modern, that have accumulated during growth, 
with one large unit. - Again in the case of a railway system, 



18 COST SEGREGATION 

the traffic may increase so that a single track bridge is incapable 
of handling the traffic, although it is still in good condition. 
In such cases the item affected has become inadequate. It may 
be argued that a larger unit should have been installed in the 
first place and in some cases that would have been right, but 
more often this would have resulted in reduced economy due 
to too early an outlay of capital. 

Only when the plan of installing smaller units first and 
replacing them with larger units later results is better economy 
than having installed the larger units at the start, have we a 
case of inadequacy of equipment. When conditions are reversed 
we have a case of inadequacy of the management. 

24. Uselessness. — A system or part thereof becomes use- 
less (without use) when the service for which it was intended is 
no longer demanded. Thus a temporary structure, used during 
construction, becomes useless and in fact often a nuisance, when 
the construction is finished. Again a logging railroad may be 
built into a certain district. When the logging is finished, the 
railroad becomes useless, though some parts of it may be 
salvaged. 

25. Fundamentally, the object of cost calculation is to 
maintain the capital intact, neither permitting it to increase 
nor decrease. For this reason, it is necessary to set aside each 
year a certain sum of money, equal to the amount that the 
system has depreciated in that time. This money, so set aside, 
forms the Depreciation Reserve or Sinking Fund, the former 
term being preferable. This depreciation is an essential part 
of the cost of the production of the service being rendered. 
The amount of the depreciation reserve laid by should be such 
that, at any time that the equipment has depreciated to zero, 
it, together with the accumulated interest thereon, should 
amount to the original cost of the equipment, less whatever 
scrap value it may have. So also if the equipment appre- 
ciates in value, this amount must be deducted from the pro- 
duction cost for the very same reasons that depreciation 
must be added. The Depreciation Reserve is not a dona- 
tion to the Servant from the employer, but a trust fund that 



LIFE 19 

must all be spent for the purposes for which it was allowed or 

returned. 

Primarily, the amount of the depreciation reserve that must 

be annually laid by depends upon the life of the structure or 

equipment. This life must be determined from experience and 

is now fairly well known. It is given as well as possible in 

the table below. 

TABLE 1 

Approximate Useful Life of Structures and Equipments 

Description LiFE,Yrs. Description Life,Yrs. 

Ash Conveyers, steam jet. ... 50 Elevators, bucket 10 

Bridges, concrete Permanent Fans, centrifugal 25 

" steel 40 Grates 10 

" wood 15 Generators, A.C 30 

Building, concrete Permanent D.C 20 

" brick 50 Heaters, feed water, closed. . . 20 

" wood or sheet-iron . . 15 " " " open..... 30 

Boilers, fire-tube 15 Hoisting Machinery 20 

" water- tube 40 Motors same as generators 

Belts, leather 8 Pumps, plunger 15 

Bins, steel 25 " centrifugal 25 

" wood 10 Producers, gas 20 

Chimneys, concrete Permanent Pd?ing 35 

" brick 50 Rotary Converters 30 

" steel, self-sustaining 35 Storage Batteries 4 

" sheet-iron 8 Switchboard 50 

Condensers, jet 20 Transformers 50 

" surface. 20 Transmission Line, pole 12 

Conveyers, bucket 20 " " steel 20 

" belt 7 Traveling Cranes 50 

Engines, high-speed 15 Turbines, steam 20 

" low-speed 25 " water 25 

Economizers 15 Wiring, electric 30 

The above list is necessarily very limited. Besides it must 
be borne in mind that the life given is necessarily rather approxi- 
mate. It depends greatly on the quality of the apparatus 
when purchased, together with a multitude of other conditions. 
The above table is for good substantial equipment, not the 
cheap stuff that is also found on the market. 

Other conditions that affect life are the surrounding ele- 
ments. Thus, the life of wood pipe is very great, if it is kept 
thoroughly impregnated with water, but deteriorates rapidly 
with disuse. So again soil conditions often are a deciding 
factor, iron pipe decomposing rapidly in alkaline soil or sea 



A 



20 COST SEGREGATION 

water. On the other hand the life of stacks is greatly influ- 
enced by the amount of sulphur in the fuel. This is due to the 
formation of sulphurous acid from the sulphur dioxide and its 
oxidation to sulphuric acid in the presence of water and free 
oxygen, when the temperature is not too high. This acid 
attacks the iron, causing very active corrosion. It seems to be 
most active during light or no load periods. So also, the life 
of boilers being fed with untreated water, depends more upon 
the chemical composition of the raw water than upon any one 
other item. 

The life of all apparatus depends upon the skill of the operat- 
ing engineers as well as the proper correlation of all parts of the 
system by the designing engineer. All such items must be 
taken into account in determining the probable life of a system. 
But the difference between normal and actual conditions should 
be charged to management. With unusually good operators a 
plant will tend to appreciate in value for quite a few years. 
This should be credited not only to the operators buj: to the 
management which had the courage and foresight to pay the 
price for first class men. On the other hand if the first cost of a 
system or the operating cost is abnormally high due to poor 
designing, the difference should be charged to bad management 
where it ultimately belongs. 

The reader is cautioned not to confuse depreciation with 
wear, as is often done in practice. Depreciation is a fixed 
charge, continuing unaltered whether the unit is in use or 
idle. Wear is an operating cost, the amount of wear being in 
direct proportion to the amount of use. When a machine is 
worn out, it has not depreciated to zero worth. It has merely 
worn out and we have merely a major repair item. 

Thus when a bearing is worn out, we replace the bearing and 
thus repair the engine. This is admittedly a repair item. 
So when an engine is worn out, we replace the engine and thus 
repair the power plant. The replacement of an entire worn- 
out engine is just as much a repair item as the replacement of 
any of its parts, the only distinction being the degree of the 
replacement. Whenever the cost of maintenance of an old 



MATERIALS CONSUMED 21 

unit is greater than the total costs (fixed charges and mainte- 
nance) of a new unit, it is evident that it will pay us to discard 
the old unit. In such a case it is worn out. 

We may therefore distinguish two lives for all things, namely 
the natural life and the operating life. The natural life 
depends on the factors of obsolescence, inadequacy and 
uselessness. The operating life depends upon the amount of 
use that may be obtained from anything before it is worn 
out. The shorter of the two determines the true life of the 
unit. 

It would be senseless and very bad engineering to make the 
operating life of a unit longer than its natural life. Yet this 
very thing has been done in the past. The large, extremely 
low speed engines of forty and fifty years ago, some of which 
are still in use, refuse to wear out, though they are hopelessly 
obsolete. It is for this reason that practically none of this 
type of machinery is now built. 

26. Taxes may be definitely determined in any community, 
usually running between one and two per cent of the assessed 
value. The assessed value is usually somewhat below the 
original cost of the undertaking and this must of course be 
taken into account. 

Insurance usually runs between 0.5 and 1.5% but this again 
depends upon the construction of the system. A concrete 
building that is absolutely fireproof need carry no insurance, 
thus tending to reduce the total cost of such a structure. If, 
for example, a building cost $100,000 and insurance is 1 %, 
then the annual insurance outlay is $1000 which, at 5%, 
interest, capitalizes at $20,000. A fireproof building would be 
worth this much more. But to all this we must add the cost 
of insurance on the equipment that is to be housed within 
the building, further increasing the value of the firepoof 
structure. 

27. Materials Consumed. — Any undertaking rendering a 
given service, whether a factory producing a given commodity 
or a power plant, uses up a certain amount of materials, which 
is either entirely consumed or else rendered into a more or less 



22 COST SEGREGATION 

worthless condition. The amount of material so consumed is 
very closely in direct proportion to the amount of service 
produced. 

Thus a steam power plant consumes fuel, lubricating oils and 
greases, waste, packing, paints, and the like. A hydro-electric 
power plant consumes all of these except fuel. A factory pro- 
ducing machinery consumes a great variety of supplies as 
well as iron, steel, brass, and the like in considerable quanti- 
ties, which are rendered into the more or less useless form of 
turnings and punchings. 

The cost of all such materials consumed should be charged 
under this head to operation, together with their cost of pur- 
chase and delivery, and the cost of waste removal, less the 
salvage if any. 

28. Attendance. — The salary or wages paid to all who are 
actually engaged in the production of a given service should 
be charged to operation under the head of attendance. In an 
iron works, for example, this includes the wages paid to machin- 
ists, molders, boiler makers, helpers, clerks, bookkeepers, col- 
lectors, and the like, together with such parts of the salary of 
the superintendent, engineer, and manager as are devoted to 
the actual production of the service. And so for other under- 
takings. 

29. Maintenance. — The cost of all materials and labor, 
engineering, and inspection service necessary to keep an under- 
taking in good running condition, so that its productive capacity 
does not decline, should be charged to operation under the 
head of maintenance. The largest item that comes under this 
head is repair and replacement of worn-out parts of equipment 
and the maintenance of grounds and buildings. But as pre- 
viously pointed out, the cost of equipment replaced because of 
obsolescence, inadequacy, or uselessness should be assessd to 
fixed charges under the head of depreciation. Attendance, 
maintenance, and materials consumed decrease with the 
increase in quality of the system, while its first cost increases 
therewith. Therefore as the fixed charges are allowed to 
increase, the operating costs decrease. This gives us one of 



STORAGE, SALE AND DISTRIBUTION 23 

our most important problems, to find, under given load con- 
ditions, where the sum of the fixed charges and operating 
costs, i.e., the total costs of production of the service, are a 
minimum. 

30. Stock Account. — It is invariably necessary in any 
undertaking, whether large or small, to carry a special, tempo- 
rary account of materials on hand for use or sale. This account 
should be divided into two parts, namely raw stock and fin- 
ished product. Thus in a steam-electric power plant, we have 
on hand usually considerable materials, such as fuel, supplies, 
and the like, which should be charged to capital until used and 
then transferred to operation. In this case we have no fin- 
ished product, for electricity is and must be used as produced, 
excepting what little is carried over in storage batteries. 
In the case of an engine factory, we have the raw materials 
account, covering materials consumed as well as those going 
into the finished product. This stands as a debit against the 
production of the engines, while the finished product stands 
as a credit account. 

31. Storage, Sale and Distribution. — As previously shown, 
the costs of rendering a given service may be divided into col- 
lection, production, transportation, and distribution, the latter 
including sales. The item of storage, sales, and distribution, 
while in some business the most important function, is no 
exception to the treatment given any of the other items in the 
complete rendition of a service. The more uniformly we 
operate throughout the year the more must be stored, i.e., we 
can decrease the production cost at the expense of the storage 
cost. To determine the conditions of best economy, in such a 
case, gives a typical problem in cost analysis. 

Selling is one of the most important services that is rendered, 
because it comprises not only the actual selling but the giving 
of advice and assistance to the purchaser in the successful use 
of the product being sold, as well as trouble shooting, that 
is the correction of troubles that appear in the use of the prod- 
uct whether due to more or less defect in the product or to 
difficulties that the purchaser gets himself into 



24 COST SEGREGATION 

32. In any actual plant all the above items of expense are 
found to be more or less mixed together since operation, main- 
tenance, replacement, and the like are taking place simul- 
taneously. They can, of course, be readily segregated by an 
engineer, not with absolute accuracy, but sufficiently so for the 
most exacting practical needs. 



CHAPTER III 
FUNDAMENTAL FINANCIAL CALCULATIONS 

Compound Interest. Principal. Equity. Term Factor. Operating 
Vestance. Depreciation Rate. Present Worth of a Depreciating 
Equipment. Depreciation Vestance. Total Vestance. Annual 
Operating Costs. 

33. Interest. — The loan of money is paid for by the annual 
payment of interest usually expressed as a certain per cent of 
the principal. The loan of money is always for a certain term 
of years only, at the end of which time the principal must be 
returned, the interest in the meantime being paid annually. 
This is so-called Simple Interest. On the other hand, instead 
of paying the interest when due, it may be added to the prin- 
cipal and become a part thereof, the sum of the two thereafter 
bearing interest. This is called Compound Interest. Thus we 
meet the problems of how much would a principal amount to 
in a certain number of years, bearing compound interest at a 
given rate, and conversely, knowing that a certain amount is 
due in a given number of years at a known rate, to find its 
present value. 

34. Compound Interest Formula. — Letting 
P = the principal, 

R = the rate of interest, 
N = the number of years, 
and A = the amount at compound interest, 

then A = P(i + R)». 

This can be easily shown as follows : 
Example i. To begin with we have 
A = P dollars. 
25 



26 FUNDAMENTAL FINANCIAL CALCULATIONS 

At the end of the first year, this has increased to " 

ili = P(i + R), 
so at the end of the second year, it is 

A 2 =P(i +R) (i +R) =P(i+R)\ 
And at the end of the third year, we have 

As = P(i + R) 2 (i + R) = P(i + R)\ 
and so on. At the end of (n) years, it is evidently 

A n = P(l +R)\ . ..' (l) 

Example 2. How much will $100 amount to in 10 years at 
8% compound interest? 
Solution: In this case 

A = ioo(i.o8) 10 
or log A = log 100 + 10 log 1.08 

= 2 + 10 X 0.0334 = 2.334, 
so that A = $215.80. 

Example 3. In how many years will a sum of money treble 
itself at 10% compound interest? 
Solution: In this case we have 
A 





p-d + 


R)», 


where 


A 

p = 3 




and 


R = 0.10, 




so that 


3 = (1.1)* 


1 


and 


n log 1.1 =log3, 




or 


n- l ° g3 
log I.] 


0.477I 
[ O.O4I4 


whence 


n = 11.52 


years. 



35. Principal. — Knowing the sum of money that will be 
due in a certain number of years at a given rate of compound 
interest, we can find its present worth, by solving the above 
formula for (P), thus 

P =VTW» (2) 



EQUITY 27 

Example 4. One thousand dollars is due in ten years, the 
compound interest rate being 7 %. What is its present worth? 

Solution: In this case 

A = $1000, 

R = 7 % 

and n = 10, 

„ 1000 
so that Jr = 



(1.07)" 

or log P = log 1000 - 10 log 1.07 

= 3 - 10 X 0.0294 = 2.706, 
whence P = $508.10. 

36. Equity. — If the operation of a certain equipment is 
guaranteed, and its actual performance falls below this, the 
cost of operation of this unit will be increased by a certain 
amount annually. In such a case a certain sum (the equity) 
must be deducted from the price of the unit to compensate the 
purchaser for this loss in guaranteed economy. On the other 
hand, if the guarantee is exceeded, a certain sum should be 
added to the price to compensate the manufacturer for the 
increased economy attained. 

In determining the equity, it is evident that if we capitalize 
this annual amount by which the guaranteed economy falls 
short (or is exceeded) at the usual rate of interest, and pay 
this sum to the purchaser, he will receive this annuity, as 
interest therefrom forever, whereas the equipment has a 
limited life only. We would thus be paying too much. Call- 
ing (a) the annuity, or the annual cost of operation in excess 
of the guarantee, and (R) the interest rate, then this capitalized 
amount (A) will equal merely 

A - -• 
A ~ R 

Instead of deducting (A) dollars from the selling price (C) 
which, as pointed out above, would be too large a sum, we 
should loan this sum (A) to the purchaser during the life of the 
unit, without charge, the principal to be returned by him at the 
end of this time, the interest thereon, would compensate the 



28 FUNDAMENTAL FINANCIAL CALCULATIONS 

purchaser for the reduced efficiency. This would be an equi- 
table adjustment. But this adjustment can be simplified. If 
instead of waiting for (n) years, the life of the equipment before 
the amount (A) is returned, we deduct from this amount (^4), 
due in (n) years, its present worth (P), then we can reach a 
complete adjustment at once. This difference (^4 - P) is the 
equity (£), so that 

E = (A- P), 
but p = (l + R y ( reL e( l- 2 ) 

and A = — 



R 



so that 



L (i + *)• J 



(i + R) n 

R L ( J + R ) n J R L (i + R) n J 

37. Alternate Solution. — We can obtain the above formula 
in a more direct but also more laborious way as follows: To 
begin with we lay aside E dollars. At the end of the first year, 
this has increased to 

E (i + R) dollars, 

from which we must pay the purchaser (a) dollars. So we 
have left 

[£(i + R)- a] dollars. 

This will increase to 

[£(i + R) - a\ (j + R) = E(i + R) 2 - a(i + R) dollars 

at the end of the second year, when we must pay out the 
second installment of the annuity. After this we have left 

£(i + R) 2 - a(i + R) - a 

which will increase to [£(i + R) 3 - o(i + R) 2 - a(i + R)~] 
at end of the third year. So then at the end of the third 
year, we have left after paying the third installment of (a) 
dollars 

£(i + R) 3 - a(i + R) 2 - a(i + R) - a, 



EQUITY 29 

and so at the end of (») years we have left 

£(1 + £)» - a(i + i?)- 1 - a(i + i?) n " 2 . . . 

a(i + £) 2 - a(i + 22) - a. 

If (») is the life of the equipment — the term of the annuity 
— then the balance left on hand should be zero, so that 

[£(1 + R)» - a(i + R)"- 1 - a(i + R) n ~ 2 . . . 

— a(i + R) — a] — o, 
or 
£(1 + *)» = a[(i + i?)"- 1 + (1 + £)- 2 . . . 

+ (1 + R) + 1]. 

We have next to find the sum of the series on the right hand. 
Calling this sum (S) , then 

£(1 + 22) n = oS, 
and S = (1 + tf)" -1 + (1 + i?) n ~ 2 • • • + (1 + -R) -+ 1. 
Multiplying through by (1 + 12), we get 
(1 + 12)5 = (1 + 22)» + (1 + R)"- 1 . . . 

+ (1 + RY + (1 + R) 2 + (1 + 22). 

Now subtracting (S) from the left hand side of this equation, and 
its value, the foregoing series, from the right hand side, we get 

(i+R)S-S = (i + R) n - 1, 

or RS = (1 + R) n - 1, 

, c (1 + R)* - 1 
and o = 7; 

But since £(1 + R) n = aS we can substitute for (5) its value 
just found and get 

E(l + *). = a p+f- 1 ], 

or 

_ fl r (i + r)* - i i _ a r 1 1 
*L (i+^) w J *L (i+*kI 

as before. 

38. Example 5. A 100 h.p. engine is guaranteed to consume 
not over 2o# of steam per h.p.h. A test shows its actual con- 
sumption to be 2 2#. If the engine is run at full load for 3000 



So FUNDAMENTAL FINANCIAL CALCULATIONS 

hours per year, and the steam costs 20 cents per iooo#, what 
will be the amount which should be deducted from the pur- 
chase price of the engine to compensate for this loss in econ- 
omy? The life of the engine is 20 years; interest rate 5 %. 

Solution: Under the above conditions, the engine evidently 
uses 2oo# of steam per hour in excess of the guarantee. In a 
year of 3000 hours it would use in excess 

3000 X 200 = 6oo,ooo#, 

600,000 vy ~ 

costing X 0.20 = $120, 

1000 

so we have that a = $120, 

R = 0.05, 

n = 20, 

whence 

„ 120 (1.05) 20 - 1 .. , 

E = / x 20 = 2 400 X 0.5161, 

0.05 (1.05) 20 

or E = $1239. 

39. Example 6. A 300 h.p. motor is sold for $1800 with a 
guaranteed efficiency of 92 %. It is used at full load for 3000 
hours per year. The life of the motor is 20 years and the power 
costs one cent per h.p.h. If the motor only develops 90% effi- 
ciency, how much should be deducted from the price to com- 
pensate the purchaser therefor, assuming an interest rate of 
6%? 

Solution: The horse power that would be consumed by the 
motor under the guarantee is 

300 -f- 0.92 = 326 h.p., 

while it actually consumes 

300 + 0.90 = 333.3 h.p., 
a difference of 

333-3 - 3 2 6 = 7-3 h.p., 

costing 7.3 cents per hour or 

0.073 X 3000 = $219 per year. 



TERM FACTOR 31 

So we have here that 

a = $219, 
R = 0.06, 
n = 20, 
so that 

219 (i.o6) 20 -i 
0.06 (1.06) 20 ' 
or 

E = $2450. 

The corrected price would then be 

1800 — 2450 = — $650. 

That is, assuming we can get a motor of 92 % efficiency from 
another firm, we could not afford to accept this motor as a 
present, unless we received therewith a cash bonus of $650. 

(1 + R) n — 1 
40. Term Factor. — The factor y —, J -—r — we shall desig- 

(1 + R) n 6 

nate for convenience the term factor ( T) , so that 

„ aT 



(4) 



In the table below, we give the values of this term factor ( T) 
for various values of the interest (R) and the life (n). 



32 FUNDAMENTAL FINANCIAL CALCULATIONS 

























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DIFFERENCE IN WORTH 33 

Example 7. What is the difference in worth per h.p. of two 
Diesel engines, one of which uses 0.40$ of oil per h.p. and the 
other uses 0.46$ oil per h.p. if the engines are used at full load 
for 4000 hours per year, the oil costing 0.5 cent per pound, 
interest 7 % and the life of the engines being 25 years? 

Solution: The difference in fuel used per h.p.h. is o.o6# and 
per year is 0.06 X 4000 = 240$, costing $1.20. So we have 







a 


= I1.20, 








R 


= 0.07, 




Whence from table 


2, 


n 


= 25 years. 






T 


= 


0.81575 




and 


E 


= 


^ X 0.81 
0.07 


575, 


or 


E 


= 


$13-9843- 





That is the engine using only 0.40$ oil per h.p.h. is worth 
about fourteen dollars more per h.p. than the engine using 0.46$. 
Evidently it would not pay to spend more than this amount 
to gain this added economy. 

Example 8. What is the difference in worth between a 
steam plant using 2# of coal per h.p.h. and a producer plant 
using i# coal per h.p.h., the coal costing $5 per ton and interest 
being 6 %, if the lives of the plants are 20 years, and they are 
run at full load for 3000 hours per year, other things being equal? 

Solution: At $5 per ton, the coal costs \ cent per pound. 
The steam plant therefor cost \ cent per h.p.h. more than the 
producer plant, or \ X 3000 = $7.50 more per year per h.p. 



So then we have 




a = $7.50, 
R = 6%, 


Whence by table 2, 




n = 20 years. 




T 


= 0.68821, 


and 


E 


- 7 * 5 ° X 0.68821, 
0.06 


or 


E 


= $86.02625. 



34 FUNDAMENTAL FINANCIAL CALCULATIONS 

That is, under the conditions assumed, the producer plant is 
worth 86 dollars more per h.p. than the steam plant. 

Example 8. Determine how much more we could pay for a 
hydro-electric power plant and transmission line, than for 
a steam plant, to be run at full load for 3000 hours per year, 
if the operating costs for the latter are 0.6 cent per h.p.h., 
while for the hydro-electric system, they are only 0.2 cent. 
Assume the life in either case to be 30 years and interest to be 
5 % while all other things are equal. 

Solution: In this case it costs 0.4 cent more per h.p.h. 
for the steam plant than for the hydro-electric plant, or 
0.004 X 3000 = $12 more per h.p. year. So we have 









a = 12.00, 










R = 5%, 




by 


table 2, 


T 


n = 30. 
= 0.76861, 








E 


12.00 w 

= X 0. 

0.05 


76861 






E 


= $184.4664. 





Example 9. A Diesel engine plant uses 0.4$ of fuel oil per 
h.p.h. costing $1.60 per barrel (320/), while the attendance 
cost is 0.1 cent per h.p.h. A steam plant uses 3# coal per 
h.p.h. costing $5 per ton with an attendance cost of 0.2 cent 
per h.p.h. These plants are assumed to have a life of 15 years. 
If interest is 5 % and the plants are to be run for 4000 hours 
per year at full load, what is the difference in worth between 
the two? 

Solution: At $1.60 per barrel, the oil costs 0.5 cent per lb., 
so that for the Diesel Engine Plant we have 

Fuel cost = 0.4 X 0.5 = 0.2 cent, 
Attendance =0.1 cent, 
Operating cost = 0.2 + 0.1 = 0.3 cent. 



OPERATING VESTANCE 35 

At $5 per ton, the coal costs 0.25 cent per lb., so that for 
the steam plant we have 

Fuel cost = 0.3 X 0.25 = 0.75 cent, 
Attendance = 0.2 cent, 
Operating cost = 0.75 + 0.2 = 0.95 cent. 

Evidently the steam plant costs 0.95 — 0.3 = 0.65 cent more 
per h.p.h. than the Diesel plant, or 0.0065 X 4000 = $26 more 
per year. So we have 

a = $26, 

n = 15 years, 
and R = 5%. 

Whence by table 2, 

T = 0.51897, 

. „ 26.00 w 

and E = X 0.51897, 

0.05 

or E = $269.8644. 

41. We have thus, as illustrated in the above problems, a 
method not only for determining the equity in any given case, 
but the same method permits us to get the difference in value 
of various units or systems only, however, under full load 
conditions. The case of relative values, and equities under 
partial load conditions, will be taken up later. We can extend 
this treatment somewhat further by determining the capital- 
ized value of the operating cost. Thus if the operating cost 
is (a) dollars per year, and since this outlay will continue for all 
times, and not merely for the life of the equipment, its total 
capitalized amount, or operating vestance ( V) as we shall call 
it, will be 

7 -i • •• - (5) 

where (R) is the rate of interest as before and (a) the annual 
operating cost. 

Example 10. If it costs $12 per year for operation to produce 
a continuous horse power and money costs us 6 %, what is the 
operating vestance (V) per h.p.? 



36 FUNDAMENTAL FINANCIAL CALCULATIONS 

Solution: Here 

a = $12, 
R = 6%, 

so that V = ^r = $200. 

o.oo 

That is by investing $200 at 6%, the proceeds thereof will pay 
the operating cost. 

42. The Depreciation Rate is a certain per cent of the first 
cost which is annually laid aside to form the depreciation 
reserve. This rate must be such that at the end of the life 
of the equipment, the depreciation reserve, together with its 
accumulated interest, will equal the first cost less the scrap 
value of the equipment. The item of accumulated interest 
on the depreciation reserve has often been overlooked in the 
past, introducing a very great error. Thus if the life of a cer- 
tain item of equipment is 50 years, it has usually been assumed 
that the depreciation rate was 2 %. We shall show later that 
this is about six times too large — an error of 600%. 

43. To solve this problem correctly, we must determine 
an amount or annuity (A) which, when laid aside each year 
for (n) years, the life of the equipment, and drawing interest 
at (R) per cent, compounded annually, will amount to (W) 
dollars, the first cost of the unit less its scrap value, i.e., its 
wearing value. The depreciation rate (D) which should be 
applied only to the wearing value (Traction Valuation Com- 
mission of Chicago, 1906) is then merely, 

D =- (6) 

It is difficult to determine for long periods in advance the scrap 
value of any item of a system, and, furthermore, this item is 
small in comparison with the first cost. In practice it is cus- 
tomary to neglect this difference and apply the depreciation 
rate to the first cost (C), getting the approximate equation 

D = A c <?> 



DEPRECIATION 37 

This will be correct if it is understood that at the time of a 
purchase of a new unit the scrap value of the replaced unit is 
applied to this purchase and not treated as an asset of the un- 
dertaking. This will give, then, exact results, when an attempt 
to determine the scrap value of a unit, 15, 20, or 30 years 
hence would be only a wild guess. The determination of the 
depreciation rate follows, as per these latter conditions. 

44. Let 

A = the depreciation annuity (or amount), 

R = the interest rate, 
C = the first cost, 
n = the life of the unit, 
and D = the depreciation rate. 

Then at the end of the first year, we deposit (^4) dollars, so 
that the depreciation reserve is 

Fi = A dollars. 

At the end of the second year, due to the interest rate, this will 
increase to A(i + R) dollars, to which we add, by deposit, 
another (A) dollars. The depreciation reserve is then 

F 2 = [4(i + R) + A~] = A[_{i + R) + 1] dollars. 

At the end of the third year this increases to [\4(i + R) + A~] 
(1 + R), to which we add another (^4) dollars by deposit, 
making the reserve then 

F 3 = A[(i + R) + 1] (1 + R) + A 
or F 3 = 4[(i + R) 2 + (1 + R) + 1], 

so at the end of the fourth year it is 

F* = Al(i + R) 3 + (1 + R) 2 + (1 + R) + 1], 

and at the end of the (n) year, the depreciation reserve is 

F n = 4[(i + R)*-i + (1 + R) n ~ 2 . . . 

+ (1 + R) 2 + (1 + R) + 1] dollars. 



38 FUNDAMENTAL FINANCIAL CALCULATIONS 

Again, calling 

S = (i + R)*- 1 + (i + R)"- 2 . . . + (i + R)* + i, 
then 

_ (i + R) n - i 

■ " R ' 

so that 

Fn = |[(l + *)" - l]. 

But at the end of (n) years, where in) is the life of the equip- 
ment, then the depreciation reserve (F„) must equal (C), the 
first cost, whence 

C=|[(H-i?)»-i], 

or 

A R 



C (i + R) n - i 
But the depreciation rate (D) equals — , by eq. (7), so that 

D= (i+i?K T i -- (8) 

Example 11. The life of an engine is 10 years, interest rate 
6%. What is the depreciation rate? 

Solution: Here w = 10 years, 
R = 0.06, 

0.0 6 
(i.o6) : 
instead of 10% usually assumed. 

Example 12. If in example 11, the interest rate is 8% 
instead of 6%, what will be the depreciation rate? 
Solution: Then n = 10 years, 
R = 8%, 

whence D = - — |^ = 6.9%. 

(i.o8) lu — 1 

The larger the interest rate or the longer the life of the 
equipment, the smaller the depreciation rate, as illustrated 
below. 



so that D = ( _ _;, 10 _ — =7.5%, 



APPRECIATION 39 

Example 13. The life of a concrete building is 50 years, 
interest rate 6%. What is the depreciation rate? 

Solution: Here R = 6%, 

n = 50 years, 

_. 0.06 _ 

so that D = (io6)5Q _ t = 0.34%. 

That is, the depreciation rate is approximately one-third of 
one per cent instead of the 2% usually assumed, giving an 
error of some 600% as mentioned before. 

In order to facilitate calculations, we give below, in Table 3, 
the depreciation rate for various values of (n) and (R), accord- 
ing to equation (8). 

45. We have spoken thus far only of depreciation. 
Machinery depreciates with age. But lands, live-stock, and 
the like almost invariably appreciate in price. Evidently 
appreciation is merely negative depreciation, so that if we 
charge depreciation as a cost, then we must charge appreciation 
as a profit. 

Lands invariably increase in value with the increase in popu- 
lation in their vicinity or in the development of their resources 
previously lying undeveloped and perhaps undiscovered. The 
increase in price of land with population is very pronounced. 
Live-stock appreciates by natural processes. But besides all 
this, all classes of things may apparently increase in price due 
in reality to a depreciation of the currency. Under such condi- 
tions, they really do not increase in worth at all. A pound of 
wheat has no more nutritive value now than a hundred years 
ago, though it now costs many times as much. Its nutritive 
value is a constant. Its increased cost is only apparent due 
to the decreased price of the money. 

So a piece of machinery may increase in price, but not worth, 
due likewise to the decreased price of gold. Where the price 
of the metals and labor has doubled, it is evident that the 
price of the machine must be doubled, which merely means, 
according to our present system of exchange, that it will take 



4Q 



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APPRECIATION 41 

twice as much gold to buy the same machine of the same worth 
as before. 

The increased production of gold results in a decrease in its 
price. The price is determined by the demand for the gold for 
such useless uses as ornamentation and exchange and for 
useful use in industry. The price that can be paid for gold for 
industrial use is strictly limited by its intrinsic worth. The 
price it may have for useless uses is a problem of mob psy- 
chology rather than one of science. As the production increases, 
the price falls, and with the fall in price, more gold is used in 
industry. When the price has fallen sufficiently, all that is pro- 
duced will be usefully used and the substance, gold or whatever 
it may be, will have fallen to its true comparative value, or 
industrial worth. 

It must be borne in mind that the prices of iron, copper and 
nickel, as well as silver and gold, went through the same stages, 
from high artificial price to real industrial worth. 

46. In the case of public utilities, depreciation is not a sum 
of money paid outright to the companies, but is in reality a 
trust fund to insure maintaining the service at par. It is the 
duty of the commissions to see that it is all properly used. 
But in taking depreciation into account, we must also take 
into account appreciation. To charge the public for that which 
depreciates without allowing for that which appreciates in 
value is eminently unjust. In private companies the matter 
should be treated likewise if the true status of the business is 
desired. 

47. While each item of a plant has usually a different depre- 
ciation rate, yet a mean rate for the entire plant may be ob- 
tained so that the plant may be treated as a whole. This is 
illustrated in the following illustrative problem. 

Example 14. A power plant of 1000 h.p. consists of the fol- 
lowing items, with costs and depreciation rates as set forth. 
Determine the mean depreciation rate. ' 

Solution: Summing up the costs gives us a total of $91,400 
for the plant. So also summing up the depreciation amounts 
gives us $1682.10 — $1280 = $402.10 net depreciation. The 



42 



FUNDAMENTAL FINANCIAL CALCULATIONS 



mean depreciation rate is evidently the net depreciation amount 
divided by the total cost, or 

402.10 -5- $91,400 = 0.44% Ans. 



Depreciation 



Item 


Cost 


Rate 


Amount 


Grounds 


$16,000.00 
9,000.00 

18,000.00 
3,800.00 

18,000.00 
3,500.00 
3,000.00 
1,800.00 
4,000 . 00 
800 . 00 
1,600.00 
2,400.00 
1,700.00 
1,000.00 
5,600.00 
1,200.00 


-8.0% 

0.3 

2.7 

4.0 

1 .0 

6.0 

0.33 

5-9 

6.0 

30 

1 .0 

o.35 
3.0 


- $1280.00 

27.OO 

486.OO 

152.00 

180.OO 


Buildings, concrete 

Turbo-generators 


Condensers 

Boilers 

Stokers 


Stacks, brick 


9.00 
106 . 20 


Coal bins, steel 


Coal conveyers 


240.OO 

24.00 

16.OO 

8.40 

CT OO 


Boiler feed and service pumps. . . . 

Feed water heaters, open 

Switchboard and wiring 


Exciters 


Foundations (mach.) 


O 00 no 


Piping and conduits. . 


3-0 
0.3 


l68 OO 


Crane. . . . . . . 


360 





48. Present Worth of a Depreciating Equipment. — As an 

equipment depreciates, the sum of its present worth plus the 
depreciation fund together with accrued interest thereon, 
equals the first cost of the equipment. It is often desirable to 
know the worth of a depreciating unit at some given time. 
This we can determine as follows: 

As before let 

C = the first cost, 
D = the depreciation rate, 
R = the interest rate, 
n = the total life of the unit, 
and m = the actual age of the unit. 

At the end of each year we lay aside DC dollars. So at the 
end of the first year the depreciation fund (F) amounts to 

F 1 = DC. 
This increases to DC(i + R) at the end of the second year, to 



WORTH OF A DEPRECIATING EQUIPMENT 43 

which we add another DC dollars. The depreciation fund 
therefore amounts to 

F 2 = DC(i + R) + DC = DC[_(i + R) + 1]. 
So also at the end of the third year the depreciation fund is 

F, = £>C[(i + R) 2 + (i + «) + 1]. 
And at the end of (m) years, it is 
F m = DC[_(i + R) m ~ l + (1 + R) m ~ 2 ■ • • + (1 + R) + 1} 

But 

(1 + R) m - 1 



(1 + R)"- 1 + (1 + R) m ~ 2 • ■ • + 1 
F m = DC 



R 

so that 

[(1 + R) m ~ 1] 



R 
But since 

D 



(l + £)»-! 

we get by substitution 

C [(l +*)»-!] 
(l +*»)-! 

The present worth (W) is the first cost. (C), less the total depre- 
ciation (Fm), so that 

W = C-F m 

which, by substitution, becomes 

r _ c _ cC(i+iO--,3 () 

(1 + #) n - 1 

Example 15. The original cost of a unit is $10,000 and its 
life is 30 years. The interest rate is 5 %. What is the worth 
of the unit at the end of the 20th year? 
Solution: In this case 

n = 3° years, 
m = 20 years, 

R = 5%, 
and C = $10,000, 

so that 

_ io,ooo[(i.o5)3° - (i.Q5) 20 1 

(i.o 5 ) 30 -i "~~ ' 
or W = $5023.90. 



44 FUNDAMENTAL FINANCIAL CALCULATIONS 

Example 16. A building has a life of 50 years. Interest is 
4%. What per cent of the first cost is its worth at an age of 
40 years? 

Solution: In this case 

n = 50 years, 
m = 40 years, 

R = 4%, 
C - 100%, 



whence 



ljr (1.04) 50 - (1.04) 4 ,_ A 

W= (i.o4)"-i ° 37-756%^. 



49. Vestance. — The first cost of the same type of equip- 
ments varies greatly. The life of the various types and their 
cost of operation offer two more intensely variable conditions. 
Besides all these we must consider the different classes of equip- 
ment that are all designed to serve the same purpose. Thus, 
for example, we have to consider the various classes of pumps, 
such as reciprocating, centrifugal, and jet pumps. And, for 
any one class, i.e., reciprocating pumps, we must consider the 
various types, such as simplex, duplex, or triplex pumps, 
single or double acting, vertical or horizontal and so forth. 

With such a kaleidoscopic mass of variable factors, it would 
seem at first hardly possible to get a basis from which compara- 
tive values of such different classes and types may be deter- 
mined. Yet such a basis may be rather readily obtained in what 
we call vestance, the equivalent cost of a permanent service. 

We shall lead up to this with the following example. 

Example 17. If we make a certain crossing with a wooden 
bridge, it will cost $10,000 and its life will be ten years. If we 
make this crossing with a steel bridge, its cost will be $18,000 
and have a life of thirty years. If the interest rate is 6 %, which 
of the two bridges will be the most economical investment? 

Solution: The first cost of the wooden bridge is $10,000. 
At the end of ten years we mus t again build the wooden bridge 
at the cost of another $10,000. Finally at the end of twenty 



VESTANCE 45 

years, we must again rebuild the wooden bridge at the cost of 
still another $10,000, when, neglecting other considerations, 
we will have attained the life of the steel bridge, 30 years, and 
rendered with the wooden bridge, or rather with three wooden 
bridges, the same service as with the steel bridge. 

To begin with, then, we invest $10,000. At the end of ten 
years we again invest another $io,ooo,whose present worth (P) is 

P A - ICXO °° - t-,Q". 

F "~ (l + R)m ~ (7^6p" $5585 ' 

So at the end of twenty years we must invest another 
$10,000, whose present worth is 

10,000 a 

$3119- 



(1.06) 



20 



The total cost of this service for thirty years, with the wooden 

bridges is 

10,000 + 5585 + 3119 = $18,704. 

The same service may be rendered with a steel bridge for 
$18,000, showing that the latter is the best by $704. Of course 
to these costs must be added that of maintenance, loss to traffic 
during reconstruction, and such other considerations as the 
physical nature of the specific problem may demand. Taking 
these into consideration, in the above case, would throw the 
decision very decidedly in favor of the steel bridge. 

50. Depreciation Vestance. — The depreciation vestance is 
the total of the present worths of the investment and reinvest- 
ments. We can determine this as follows: 

Let C = the first cost, 

n = the life in years, 
and R = the interest rate. 

To begin with, we invest (C) dollars. At the end of (n) 
years, we must invest another (C) dollars, whose present worth is 

p - c 



P 2 n = 



46 FUNDAMENTAL FINANCIAL CALCULATIONS 

Again at the end of {271) years, we must invest another (p) 
dollars, whose present worth is 

C 

(1 + R) 2 " 

and so again at the end of (3^) years, we must invest another 
(C) dollars, whose present worth is 

P - C 

3n (1 + R)*» 

and so on. So in general at the end of (mn) years we must 
again, for the wth time, invest (C) dollars, with a present 
worth of 

P - C 

mn (1 + R)™ 

Summing all these up gives 

c c c c 

Vd = C+ (i + 22)» + (i + R) 2n + (i+*) 8 » ' ' • + {i+Sp > 
or 

Let (5) equal the sum of the series in the brackets. Then 
5 - I+7 _i^ +7 _I^ r . . .- + 



(1 + R) n (1 + R) 2n ' (1 + R) mn 

Reducing to a common denominator, gives 

(i+R) mn +(i+R) {m - 1)n +(i+R) (m - 2)n ■ • .+(i+i?) 2re +(i+Jg) rt +i , , 

a- (1 + *)»» ' w 

Now multiply through by (1 + i?) n and we get 

d+«-5- d+ier» ■ (b) 

And subtracting equation (a) from equation (b) gives 
and 

= ( I + ^)U+Dn_ I 

= (l + U)»»[(l + £)» ~ I J 



DEPRECIATION VESTANCE 47 

Now dividing through by (1 + R) mn gives 

(l + Rym+Dn-mn _ ^__ ( x + £), 



(1 + R) mn = (1 + R) mn 

(l + £)" - I " (l + R) n ~ I 

When (ww), the length of the service becomes infinite, then 

(1 + R) mn also becomes infinite and -. -r — becomes zero. 

. (1 + R) mn 

So for the complete summation (m = 00 ) of the above series, 
we get 

(1 + R)* 
(1 + R)» - 1 
And since 

V d = CS 
we have 

_, C(i + R)" , . 

Fd -(i+^-i (I0) 

(1 + R) n — 1 

But we had that - — 7 -—, is the term factor ( T\, so that 

(1 +R) n 

V d =^ .(11) 

Example 18. A concrete building has a first cost of $100,000 
and a life of 50 years. If the interest rate is 5%, what is the 
depreciation vestance? 

Solution: In this case 

C = $100,000, 

R = 5%> 
and n = 50 years, 

so that by Table 2 

T = 0.91279, 

and the depreciation vestance is 

rr 100,000 ^ . 

V d = = $109,^0 A ns. 

0.91279 

If the interest rate were only 3 %, then the term factor would 
be, by Table 2, only 0.77189 and the depreciation vestance 
would be 

T7 100,000 * 

Vd = r,.L^ = $ I2 9>54o. 
0.77109 



48, FUNDAMENTAL FINANCIAL CALCULATIONS 

Example 19. A wooden building to take the place of the 
concrete building of problem (18), would cost $70,000 and have 
a life of 20 years. With interest at 5 %, which would be the best 
investment? 



Solution: Here 

so that 
and 



C = $70,000, 
n = 20 years, 

R = 5%, 

T = 0.62312, 



V d = -i-i = $112,340. 

0.62312 

This would make a difference of 

112,340 - 109,550 = $2790 

in favor of the concrete building, even excluding the higher 
maintenance cost of the cheaper structure. 

If the interest rate were 3%, then the term 'factor (T) 
would be 0.44632 and the depreciation vestance would be 

T7 70,000 = $I5M38< 



O.44632. 

In this case the difference would be 

156,838 — 129,540 = $27,298. 

Example 20. A permanent concrete viaduct costs $5000. 
What is its depreciation vestance? 

Solution: In this case n = 00 so that the term factor is 
unity. The depreciation vestance is then 

Vd = 552^ = f sooo , 

I 

the same as the first cost. 

51. Operating Vestance. — With depreciation vestance, we 
get a method of comparing equipments having different first 
costs and different lives. We thus have a method of determin- 
ing comparative values as to fixed charges and those only. 



OPERATING VESTANCE 49 

The operating costs are continuous, and are not limited to a 
term of years. By capitalizing the annual cost of operation, 
we get, therefore, the operating vestance. 
Letting 

a = operating cost per year, and 
R = interest rate, 
then 

Va = I • • • .' (12) 

This is on the assumption that the operating cost is constant. 
However, improvement in design is what renders equipments 
obsolete, and this refers particularly to reductions in the cost of 
operating. This more than compensates for obsolescence and 
both it and reduction in cost of operation must be taken into 
consideration. 

52. Total Vestance. — The total vestance (V) is evidently 
the sum of the depreciation and operating vestance, so we have 

V = V d + V a , 

V =f + l fa) 



or C 



= T ( V -i) <'3A) 



So that knowing the life, operating cost and vestance, we can 
determine what the cost should be. Also since 



: , (13B) 



VR-a ' 

we can determine the life of the unit, knowing the first cost, 
vestance and operating cost. In a similar manner we can solve 
for any of the other quantities in equation (13). 

Example 21. A 1000 h.p. Diesel Engine plant costs $75,000. 
The life of the engine is 20 years. The interest rate is 5 %. 
The cost of operation is 0.4 cent per horse power hour (h.p.h.). 



So FUNDAMENTAL FINANCIAL CALCULATIONS 

If the engine is operated at full load for 3000 hours per year, 
what will be the vestances? 

Solution: Here 

C = $75>°°°, 

R = 5%, 

n = 20 years, 



so that 
and 



T = 0.62312, 

Tr 7^,000 ^ 

V d =- L f = $120,360. 

0.62312 



The total annual operating cost is 

1000 X 0.004 X 3000 = $12,000, 
so that the operating vestance is 

12,000 a 

V a = l = $240,000. 

O.O5 

The total vestance is therefore 

120,360 + 240,000 = $360,360. 

Example 22. If, in Example 20, we put in a steam plant, the 
first cost will be $50,000, and the life is likewise 20 years, while 
the cost of operation will be 0.7 cent per h.p.h. With the 
same interest rate, i.e., 5%, which will be the best investment? 

Solution: In this case 

Vd = j^ _ $g0)24cx 

0.62312 
The annual cost of operation is 

1000 X .007 X 3000 = $21,000. 
The operating vestance therefore is 
T7 21 ,000 ft 

V a = = $420,000, 

0.05 

giving a total vestance of 

420,000 + 80,240 = $500,240. 



TOTAL VESTANCE 51 

The difference between the total vestance of the steam and 
Diesel plant is 

500,240 - 360,360 = $139,880 

in favor of the Diesel plant. 

Example 23. If the power were purchased at one cent per 
h.p.h. and the motor cost $6000 and had 90% efficiency with a 
life of 30 years, how would the total vestance compare with 
that of the steam and Diesel plants? 

Solution: In this case 

n = 3° years, 
C = $6000, 

R = 5%, 



so that 



T = 0.76861 



and V d = -*£f£- = $7810. 

O.76861 

The total annual cost of operation is 

1000 X °- PI X 3 °°° = $33,333 

O.9O 

and the operating vestance is 

V a = ^-^ = $666,667 
O.O5 

giving a total vestance of 

7810 + 666,667 = $674,477- 
This gives a difference of 

674,477 ~ 500,240 = $174,237 
in favor of the steam plant, and a difference of 

674,477 ~ 360,360= $314,117 

in favor of the Diesel plant. 

Example 24. At what price would the electric power in 
Example 23 have to be purchased in order that the electric 
motor plant may equal (a) the steam plant of Example 22 
and (b) the Diesel plant of Example 21? 



52 FUNDAMENTAL FINANCIAL CALCULATIONS 

Solution: Let p = the price of the electric power per h.p.h. 
Then the annual cost of operation is 

iooo X p X 3000 

Q = 3,333,333£; 

the operating vestance is 

Va = 3,333,3331 = $ 66) 666,657£ 
0.05 

and the total vestance is 

V = $66,666,667^ + $7810. 

(a) If the total vestance of the motor and steam plant are 
equal, then 

66,666,667^ + 7810 = $500,240, 
so that 

p = $0.0074 

or a little over 0.7 of a cent. 

(b) If the total vestance of the motor and Diesel plants are 
made equal then 

66,666,667^+ 7810= $360,360 
so that 

p = $0.0053 

or a little over J cent per h.p.h. 
53. Annual Operating Cost. — If we call 

L = the load in h.p., 

a = the operating cost per h.p.h. 
and 

N = the hours of operation per year, 
then, if (L) and (a) are constant, the total operating cost per 
year (^4) is 

A = LaN (14) 

But if the load and operating cost vary, the annual cost of 

operation will be 

/8760 
La dN (15) 

the upper limit (8760) being the total number of hours in the 



ANNUAL OPERATING COST 53 

year and therefore the maximum number of hours per year 
that any unit may operate. 

Example 25. For a certain power plant, we find that the load 
variation with the hours of operation may be expressed by the 
following approximate equation 

L = 7000+ 0.2N — 0.0001A 7 " 2 , 

while the operating cost per h.p.h. varies with the load in 
accordance with the following approximate equation 

a = 0.008 — 0.0000004Z,. 
What will be the total annual operating cost? 

Solution: Substituting in the equation for (a), the value of 
(Z,) in terms of (N) we get 

a = 0.008 — 0.0000004 (7000 + 0.2N — o. 0001 N 2 ) 
or 

a = 0.0052 — 0.00000008./V + o. 00000000004 A^ 2 
whence 

La = (7000 + 0.2N — o.oooiA^ 2 ) (0.0052— o.ooooooo8A" 

+ 0.00000000004 N 2 ) , 
or 

La = (36.40 + 0.00487V — 0.00000024A 7 " 2 + 0.0000000000 i6A" 3 

— 0.000000000000004 N 4 ). 
Substituting this in equation (15), we get 



<-r 



760 

LadN 



so that in this case 

A = 

[36.40N + 0.00024A" 2 — o.ooooooo8o53A" 3 

+ 0.000000000004 iV 4 + 0.000000000000000 N^ 60 
or 

A = 318,864 + 18,417.26 - 54,134.69 + 23,544.94 - 41,268.26, 

so that 

A = $265,433.25 Ans. 

54. In any actual problem in practice, the load assumed is 
that which is actually carried, according to the records, or that 
which, in the light of previous experience, may be anticipated. 



54 FUNDAMENTAL FINANCIAL CALCULATIONS 

The curve for the variation of the operating cost with the 
load is obtained either from previous experience or from the 
factory guarantee. 

In the examples given, it appears that the operating vestance 
far exceeds the depreciation vestance. Particularly is this true 
of equipments with low first cost. We can materially reduce 
the total vestance in many systems and therefore the produc- 
tion cost by putting in equipment with higher efficiency and 
allowing a greater first cost than we are accustomed to. How- 
ever, as we shall see, conditions are sometimes just the 
opposite, i.e., in cases we sometimes allow far too much for first 
cost. 

So far in vestance we have taken into consideration only 
interest and depreciation. There are, however, the two other 
factors of fixed charges, namely taxes and insurance. These 
factors run continuously, i.e., they are not limited to the life 
of the equipment. Their vestance may, therefore, be found by 
simply capitalizing the amount of tax and insurance. 

Letting / = the amount of insurance per year, 
X= the amount of taxes per year, 
R = the interest rate, 



then 



Vi = the insurance vestance, 
V x = the tax vestance, 



v - x 



and v i — -«• 

The total vestance is then 

V = V d + V x + Vi+ V a 

or V = -f+R+R+R- f + R 

Example 26. Determine the total vestance of a wooden 
building under the following conditions: First cost $10,000, 



ANNUAL OPERATING COST 55 

life 20 years, interest rate 7%, taxes 0.8%, insurance 1 % 
and maintenance and repairs $120 per year. 
Solution: The term factor under these conditions is 0.74158 

so that 

Jr 10,000 # 
V d = — : — 5- = $13,200, 
0.74158 

X = 10,000 X 0.008 = 80, 

/ = 10,000 X 0.01 = 100, 

and A = 120 

Total =$300 

VlAX = ^ = $4286. 
O.O7 

The total vestance is then 

V = 13,200 + 4286 = $17,486. 

Taxes affect all items of equal value alike, but the insurance 
varies with the hazard. An isolated concrete structure, 
having within it only incombustible material, need carry no 
insurance. But a wooden building in a district with much fire 
hazard must bear a heavy insurance rate, due both to itself 
and its surroundings, reducing greatly its comparative value. 

In a power plant, we almost invariably have fireproof equip- 
ment, such as engines, boilers, feed- water heaters, pumps, etc., 
housed in a building which may not itself be fireproof. If the 
structure burns down, this equipment would be destroyed. 
It is, therefore, necessary to carry insurance on this machinery. 
The question arises as to whether the cost of insurance of 
this equipment should be charged against the equipment or 
against the building. Again we may have patterns stored in 
a building which is itself quite fireproof. The patterns are 
highly combustible. If they burn they usually destroy the 
building. 

In any such case it is evident that the combustible items 
should be charged with the insurance on all that might be 
destroyed with the burning of these items. In the case of the 
power plant, the insurance on the machinery housed should be 
charged against the building, while in the second instance 



56 FUNDAMENTAL FINANCIAL CALCULATIONS 

cited above, the insurance on the building should be charged 
against the patterns stored. But that part of the insurance 
which is collected because of conflagration hazard should be 
charged against the location, i.e., the lot itself, because this 
cost is entirely due to its location. 

This much is definite. But how should the insurance cost be 
apportioned where both the building and the things stored are 
combustible? This does not lead to a definite solution. The 
charge should be apportioned to the hazard cause by each, 
i.e., the structure and the items housed and this, for want of 
sufficient data, remains a matter of judgment based on such 
experience as we now have. 

» 
PROBLEMS 

1. How much will $i amount to in 500 years at 4% compound interest? 

2. What is the present worth of $5000 due in 20 years? 

3. In how many years will $1000 quadruple itself at 8% interest? 

4. A building is sold on the following terms: The selling price of the 
building is $4000 and is to bear interest at the rate of 7% on all unpaid 
amounts. The purchaser is to pay at the rate of $50 per month. From 
this amount interest is paid and all above interest is to apply towards a 
reduction of capital due. 

(a) In how many years will the purchaser complete payment? (b) How 
much money will the purchaser have paid the seller when payment is com- 
pleted? (c) If the building depreciates at the rate of 7% per year, how 
much is the building worth when payment is completed? 

5. A ten h.p. induction motor of 87% efficiency cost $157. How much 
could we afford to pay for a motor of 88.5% efficiency if the power costs 
2 cents per h.p.h. and the motor is to be run at full load for 3000 hours 
per year? Assume the life of the motor as 25 years and interest at 7%. 

6. A 1000 h.p. Diesel engine is guaranteed to operate on 0.40$ of fuel 
oil per h.p.h. The engine operates at full load 4000 hours per year. The 
cost of the engine is $52,000. Interest is 5%. Fuel costs $1.60 per barrel 
of 32o# and the life of the engine is 20 years, while it actually uses 0.43$ 
of fuel oil per h.p.h. How much should be deducted from the purchase 
price to compensate the purchaser for this loss in economy? 

7. A 150 h.p. 220 volt motor costs $1170 and is guaranteed at 91% 
efficiency. The motor has a life of 25 years and is to be used for 2500 
hours per year. Interest rate is 6% and power costs 1 cent per h.p.h. 
If instead we purchased at the same price and other conditions a 440 volt 
motor with only 90 % efficiency, how much would we lose thereby? 



PROBLEMS 57 

8. A storage battery costs $75 per kw.h. of capacity. Its life is assumed 
at 3 years, its efficiency at 65%, other operating costs zero. Interest rate 
6%. What is the total vestance of the storage battery per kw.h., if it is 
used at full load for 2000 hours per year, and electric power costs 1 cent per 
kw.h.? 

9. A permanent concrete viaduct costs $60,000. With interest at 7%, 
determine the total vestance of this structure. What would be the vestance 
if the interest rate was 4%? 

10. Calculate the depreciation rate of an engine whose life is 21 years 
with interest at 6.5%. 

11. Calculate the depreciation rate of a stack whose life is 42 years with 
interest rate at 4.5%. 

12. An electric motor costs $300 and has a life of 30 years. Determine 
its worth at the end of o, 5, 10, 15, 20, 25, and 30 years. Assume an interest 
rate of 6%. Plot the curve. 

13. A Diesel engine cost $75,000 and has a life of 15 years. Determine 
its worth at the end of o, 3, 6, 9, 12, and 15 years. Assume 5% interest 
rate. Plot the curve. 

14. Which would be the best investment, a wooden building whose life 
is ten years or a steel-brick building, costing twice as much, and having a 
life of 40 years? Assume interest 5%, and operating costs zero. 

15. In problem (14) what would have to be the life of the wooden build- 
ing in order to equal the steel-brick building in worth? 

16. A steam plant has a first cost of $70,000 and a guaranteed life of 
25 years. The cost of attendance is 0.4 cent per h.p.h., maintenance 
0.1 cent, and repair 0.1 cent per h.p.h. With interest rate 6%, what will 
be the vestances of this plant if it is used for 4000 hours per year at full 
load? 

17. To operate a certain pump at full load for 1500 hours per year we 
can use a 10 h.p. gas engine costing $500 and having a life of 20 years. 
The operating costs for this engine are one cent per h.p.h. An electric 
motor to take the place of this engine would cost only $150 and have a 
guaranteed life of 30 years, and an efficiency of 88%. The power company 
offers power at 2 cents per h.p.h. Would the gas engine or motor be the 
best investment under these conditions, assuming other operating costs as 
zero? What would be the difference in worth? 

18. If the electric power is offered at one cent per h.p.h. in problem (17), 
then which would be the best investment? What then would be their 
differences in worth? > * 

19. In problem (17), how much could you pay the power company so 
that the comparative value (vestance) of the two propositions will be 
equal? 

20. In problem (17), if the pump was to be operated only 1000 hours 
per year which would be the best investment? 



58 FUNDAMENTAL FINANCIAL CALCULATIONS 

21. In problem (17), if the pump was to be operated only 500 hours per 
year, which of the two plants would be the best investment? 

22. In problem (17), for how many hours of operation would the com- 
parative value of the two plants be equal? 

23. For a certain plant, the load in h.p. varies according to the follow- 
ing equation: 

L = 12,000 — 0.3N, 

and the operating cost per h.p.h. varies according to the equation 
a = 0.007 — o.oooooo5iV. 

What will be the total annual cost of operation? 

24. For a certain plant the load in h.p. varies according to the equation 

L = 500 + o.22V\ 
and the operating cost per h.p.h. varies according to the equation 
a = 0.004 — o.oooooo2iV. 

What will be the total annual cost of operation? 

25. A structure costs $60,000 and has 15 years life. Interest rate is 
4%, insurance 1%, taxes 0.7%, cost of maintenance and repair 1.2%. 
Determine the total vestance of this structure. 

26. In problem (25) how much could we afford to pay for a permanent 
fireproof structure whose cost of maintenance and repair was 0.5%? 



CHAPTER IV 
BASIC COSTS 

First and Operating Costs at FuJl and Fractional Loads of Steam Engines, 
Boilers, Buildings, Centrifugal Pumps, Induction Motors, Direct 
Current Motors, Generators, Transformers, Oil Engines, Producer 
Gas Engines, Diesel Engines, Producers, Standard Pipe, Casing, Wood 
and Riveted Steel Pipe. Cost of Tunnels, Canals, Excavations, 
Hydro-electric Installations, Compressors, Condensors, Economizers, 
Fans, Feed Water Heaters, Stokers, Water Treating Plants, Super- 
heaters and Turbines. 

55. Basic Costs. — Every science, in its practical applica- 
tion, is based upon experimental data. These data are not 
absolutely exact but usually are exact enough to give good 
working results. In electricity, for example, we use in our cal- 
culations the experimental values of such constants as the co- 
efficients of resistance, inductance, and the like. The theory 
is absolutely correct. But these constants are only closely 
approximate. In heat we have such constants as the specific 
heat of the various substances, latent heats, fuel values, and 
the like, determined by experiment. So in financial engineer- 
ing we have for our "experimental" data, the basic costs of our 
equipment, structures, and the like, of materials consumed, 
such as fuel, oil, iron, and the like and finally of our costs of 
attendance and maintenance. We must know how these costs 
vary with differences in size, class, and type of the equipments, 
and the like. These data will only be closely approximate. 
But as their importance in engineering is better understood, 
more effort will be expended in getting still better data. Cost 
data are, however, more uniform and accurate than other 
scientific data. 

It seems peculiar and is indeed very unfortunate that so 
many authors in their engineering books give no, or very little, 

59 



60 BASIC COSTS 

consideration to costs, in spite of the fact that the primary 
duty of the engineer is to consider costs in order to attain real 
economy — to get the most power, for example, not from the 
least number of pounds of steam, but from the least possible 
number of dollars and cents: to get the best financial efficiency. 
Two very notable exceptions to the above are Fernold and 
Orrok's ''Engineering of Power Plants " in which a very great 
deal of valuable engineering data is given, and Harding and 
Willard's "Mechanical Equipment of Buildings." 1 

56. Records. — The object of keeping records is to furnish 
data for the engineering staff from which they may devise 
means of improving the economy of the system, and of predict- 
ing the costs of future operations. But, in many organizations, 
the recording or so-called accounting department is made 
independent of, instead of subsidiary, to the engineering divi- 
sion. Records are of absolutely no value unless they are studied 
and conclusions drawn therefrom that will aid in getting still 
further improved economy. This is well stated by Gebhardt * 
in " Steam Power Plant Engineering" when he says: "Few 
engineers realize the importance of a detailed system of account- 
ing, or the saving which may be effected in cost of operation by 
careful study of the daily records of performance. Many 
regard graphical load curves, meter readings, and similar records 
as interesting, but of little economic value. During the past 
few years, the author has made a close study of the cost of 
power in a large number of central and isolated stations in 
Chicago, and found, without exception, that the highest 
economy was effected by the engineers who kept the most 
systematic records; the poorest results were obtained where 
records were kept indifferently or not at all. In some small 
plants the numerous duties of the engineer prevented him from 
devoting the necessary time, but in the majority of cases the 
absence of records was due entirely to a lack of interest. Power 
plant records to be of value must be closely studied with a view 
to improvement. The mere accumulation of data to be filed 
away and never again referred to is a waste of time and money." 
1 John Wiley & Sons, Inc. 



RECORDS 61 

Where, however, the bookkeeping is independently done, 
instead of under the supervision of the engineers, it is usually 
unavailable both as to form and location. 

57. In basic costs, we have to consider the variation of first 
costs of different equipment with size, the variation of attend- 
ance and economy in materials consumed with load, as well as 
the interrelation of one part of a system with another as to 
costs. Thus, if we do not put in condensers in a steam plant, 
we will reduce our first costs by this amount. But at the same 
time, due to the increased steam consumption of the engines, 
it will be necessary to increase the size of the boilers, steam 
pipe, and the like, often by some 30 or 40%; the increased cost 
of these items may far more than offset the cost of the con- 
densers. So also in a pumping system, a reduction in the size 
of the pipe will make it necessary to install. a larger motor or 
engine to drive the pump, due to increased friction head, 
increasing not only the first cost of this part of the system, 
but the operating cost as well. And so for almost any other 
item of equipment. In thus attempting to "save " money, care 
must be used that the cost of the saving does not exceed the 
amount saved. 

It is well known that there is an enormous difference in the 
cost of apparatus for rendering a given service. A jet pump 
will deliver water just as a centrifugal or plunger pump will. 
And it usually costs only 2 or 3 % ?s much. Yet the difference 
in economy more than warrants purchasing the more expensive 
types. The cost of steam engines varies from $7 per h.p. for a 
simple, high-speed engine to $40 per h.p. for a horizontal 
four- valve engine. The question naturally arises just how much 
can be paid for a given unit. It is by no means true that the 
most expensive is the best. The test is of course a comparison 
of the total vestance of both equipments for anticipated load 
conditions. 

Off-hand the giving of prices seems a rather hopeless under- 
taking. There is first the variation in price of the same article 
of practically equal quality as between various manufacturers, 
due to their lack of knowledge of exact unit costs; the varia- 



62 BASIC COSTS 

tion of price as between different localities, and the variation 
of price with speed, type, and class of equipment. Yet, if in 
choosing cost for our curves, we confine ourselves in speed, 
type, and so forth, to standard practice, as we have done in the 
following curves, the matter simplifies itself greatly. 

Again there is a continual variation in price with time. 
Besides short fluctuations which may be either up or down, 
there is a continual tendency for the price to increase with 
time as measured by the weight of gold necessary to effect a 
purchase. This is due to the continual increase in the gold 
reserve, due to the fact that more gold is annually produced 
than is used in the industries. There must therefore be a 
decrease in the worth of the gold and thus an increase in price 
of commodities. Besides this, there are price variations due to 
extraordinary conditions as of war. However, the worth of the 
units, or their comparative values do not change except where 
further improvements are made or new properties and uses 
are found and developed. With the exception of the last, a 
change of price does not change the size of a one h.p. motor, 
nor is the food value of a pound of wheat altered though the 
price doubles. 

In any event, the following data are intended primarily as a 
first approximation and as illustrative of the method of applica- 
tion of the technique of financial engineering to concrete 
examples. In any specific case, exact data may be readily 
obtained. The prices given below may be converted to prices 
as of today, by dividing them by the comparative value of the 
dollar. 

58. Steam Engines. — The average price of the principal 
types of steam engines is given below. 

TABLE 4 

Simple High Speed Engines — Erected 

Average 1906-10 Prices 

Non-condensing 

I.h.p 20 30 40 50 75 100 150 200 250 

Price per i. h.p. .$34.75 $25.60 $21.00 $18.25 $14.60 $12.75 $10.90 $10.00 $9.40 



STEAM ENGINES 63 

The equation for these values is approximately 

P = ^ + 7-2 5 (16) 

or C = 550 + 7.25M (17) 

where P = the price per i.h.p., 
and M = the rated size, in i.h.p., 
C = the price of the engine. 

TABLE 5 
Simple Corliss Engines — Erected 
1906-10 PRICES 
Non-condensing 

I.h.p 100 125 150 175 200 250 300 350 

Price per i.h.p. ... -$2 1. 50 $20.00 $19.00 $18.25 $17.75 $17.00 $16.50 $16.15 

I.h.p 400 450 500 

Price per i.h.p $15 .87 $15.67 $15.50 



The equation for these values is 

C = 750 + 14M (19) 



P = ^ + 14 (18) 



TABLE 6 

Compound Engines — Erected 

1906-10 PRICES 

Condensing 

I.h.p 100 200 300 400 500 600 700 

Price per i.h.p $33.00 $25.00 $22.33 $21.00 $20.25 $19-67 $19.43 

I.h.p 800 900 1000 1500 2000 

Price per i.h.p $19.00 $18.78 $18.60 $18.07 $17 -8o 

The equation for these values is 

r, l0 °O . ( n 

p " ~w + I7 (2o) 

C = 1600 + 17M (21) 



6 4 



BASIC COSTS 



TABLE 7 
Turbo-Generators 
1906-10 PRICES 
Direct connected a. c. 60 cycle, condensing 



Kv.a 5°° 75° 

Price per kv.a $21 .90 $18. 23 

Kv.a 5000 

Price per kv.a $12 .00 

The equation for these values is 

C = 55.00 + 10.90M 



1000 
$16.40 

7500 
$11.63 



2000 

$I3-6S 

10,000 

$11-45 



3000 
$12.73 



(22) 
(23) 



$3b 


| 








































30 




\ 








































\ 






































25 




\ 


j^Co 


npou 


adC 


mdei 


sing 


Eng 


nes 
































































J20 










s 


X 






































































1 ; 15 


\ 






























'Dire 


ctCo 
bo-ge 


nnec 
nera 


ed 




\ 










-SimW 


OD-CC 

i Eng 


nden 
ines 


sing 


















O 


c 


brlis 










10 




V 










Ere 


cted 






























t Si 


nple 


Higl 
Er 


rspet 

;cted 


d En 


glne< 


























5 





























































































































200 



400 



600 



800 



1000 
I.H.P. 



1200 1400 1600 1800 2000 



Fig. i. — Showing the variation in cost per i.h.p. of steam engines and 
turbines with size and type. 

59. Operating Costs. — The number of pounds of dry sat- 
urated steam used per indicated horse power hour (i.h.p.h.) 
for compound condensing engines with 26-inch vacuum and 
ioo# gage initial pressure is as given in table 8 : 



OPERATING COSTS 65 

TABLE 8 
Steam Consumption 

COMPOUND CONDENSING ENGINES 

I.h.p 50 60 70 80 100 125 150 200 250 

Lbs. peri.h.p $20.2 $19.6 $19.1 $18.7 $18.0 $17.5 $17.3 $16.7 $16.4 

I.h.p 300 400 500 600 700 800 900 1000 

Lbs.peri.h.p $16.1 $15.7 $15.4 $15.2 $15.1 $15.0 $14.9 $14.8 

These values correspond to the equation 

w - 71 + I3 - 2 ' (24) 

where W = the pounds of steam per i.h.p.h., 
and M = the size in i.h.p. 

With fractional loads the steam consumption is as given in 
the following tables. In these tables, only the strokes of the 
engines are given. Since the allowable piston speed is well 
known, the stroke determines the r.p.m. The i.h.p. for any of 
these engines may therefore be readily determined, for any 
given bore, which may be anything from one-half to one times 
the stroke. A stroke equal to ij times the bore is most com- 
monly used, but good practice allows considerable deviation. 
Throughout, all data are given for well-designed and well- 
constructed machinery and not the cheap equipment that is 
made to sell, not to use. 



66 



BASIC COSTS 



TABLE g 

Steam Consumption 
simple high speed engines — noncondensing 



Stroke 



14 



18" 



Pressure 



8o-go# 


IOO-IIO# 


I20-I30# 


I40-I50# 


30.O 


29.6 


29 -3 


29.O 


30.6 


29.8 


29. 1 


28. s 


34-o 


32.2 


305 


29. 1 


40.0 


38.4 


370 


35-6 


30.8 


30.5 


30.2 


30.0 


29.6 


29. 2 


28.9 


28.6 


30.2 


29.4 


28.7 


28.1 


33-o 


314 


30.0 


28.6 


38.7 


37-2 


36.0 


34-8 


30.3 


30.0 


29.7 


29 5 


29. 1 


28.7 


28.4 


28.1 


29.7 


28.9 


28.2 


27.6 


32.0 


30.6 


29 -3 


28.1 


37-5 


36.1 


350 


34o 


29.8 


29 -5 


29. 2 


29.0 


28.5 


28.1 


27.8 


27-5 


29. 2 


28.4 


27.7 


27.1 


31.0 


29.8 


28.6 


27-5 


36.2 


35o 


340 


33 -o 


29 -3 


29.0 


28.7 


28. s 


27.9 


27-5 


27.2 


26.9 


28.5 


27.8 


27.1 


26.5 


30.0 


29.0 


28.0 


27.0 


35-o 


34-0 


33-0 


32.0 


28.8 


28.5 


28.2 


28.0 



STEAM CONSUMPTION 



67 



TABLE 10 

Steam Consumption 
high speed tandem compound engines 











Pressure 




Stroke 


Load 












ioo-iio# 


120-130! 


I40-I50# 


ioo-iio# 


I2O-I30# 


i40-i5o# 




4 


Condensing 26" vacuum 


Noncondensing 


IO-12" 


19-3 


18. 5 


17.7 


25-5 


24.7 


24.O 




3 

4 


20.I 


19-3 


18.6 


28.O 


27. 2 


26.4 




1 


22.0 


21 .0 


20.4 


32.6 


31.6 


30.6 




1 
4 


26.5 


26.0 


25-5 


44.O 


43-0 


42.O 




li 


20.8 


19.6 


19.0 


27.O 


25-7 


255 


14" 


4 


18.8 


18.0 


17.2 


24.8 


24.0 


23-3 




3 
4 


19-5 


18.7 


18.0 


27.O 


26.2 


25 5 




2 

4 


21 .2 


20.4 


19.7 


32.O 


31.0 


30.0 




1 
4 


26.O 


25 5 


25.0 


43-6 


42.6 


41.6 




ll 


I9.9 


18.8 


18.2 


26.0 


25.0 


24-5 


16" 


! 


18.2 


17-4 


16.7 


24.0 


23.2 


22.5 




3 

4 


18.9 


18. 1 


17.4 


26.0 


25.2 


245 




s 

4 


2O.4 


19.8 


19.0 


31-5 


30-5 


29-5 




1 

4 


25 -5 


25.0 


24-5 


43-3 


42.3 


4i-3 




ll 


19. I 


18.3 


17-5 


25.0 


24. 2 


23-5 


18* 


4 

4 


17-5 


16.7 


16.0 


23.0 


22.2 


21.5 




3 

4 


18.2 


17-4 


16.7 


25.0 


24.2 


23-5 




2 
4 


19-5 


19.2 


18.3 


31.0 


30.0 


29.0 




1 

4 


25.O 


24-5 


24.0 


43 -o 


42.0 


41.0 




i£ 


18.3 


17-5 


16.8 


24.0 


23-4 


22.5 



68 



BASIC COSTS 



TABLE ii 

Steam Consumption 
simple noncondensing pour valve engines 













Pressure 








Stroke 


LOAD 
















ioo-5# 


iio-5# 


I20-5# 


i30-5# 


140-5/ 


i5o-5# 


i6o-5# 


18" 


| 


26.O 


25-7 


254 


25-2 


24.9 


24.7 


24-5 




4 


27.O 


26.3 


25.6 


25.O 


24-3 


23.8 


23-3 




f 


29.7 


28.5 


27-5 


26.5 


25-5 


24.7 


23-9 




1 
4 


35-5 


345 


33-6 


32.7 


31-7 


31.O 


30.3 




ii 


27.1 


26.8 


26.5 


26.2 


26.O 


25-7 


25-5 


2l" 


4 


25-8 


25-5 


25.2 


25.O 


24.7 


24-5 


24-3 




4 


26.8 


26. 1 


25-4 


24.8 


24.I 


23.6 


23.1 




2 

4 


29-5 


28.3 


27-3 


26.3 


25-3 


24-5 


23-7 




1 
4 


3S-o 


34o 


33- 1 


32.2 


31.2 


30.5 


29.8 




ii 


26.9 


26.6 


26.3 


26.O 


25.8 


25 -5 


25-3 


24" 


I 


25.6 


25-3 


25.0 


24.8 


24-5 


24-3 


24.1 




3 
4 


26.6 


25-9 


25.2 


24.6 


23-9 


23-4 


22.9 




f 


29-3 


28.1 


27.1 


26. I 


25- 1 


24-3 


23 5 




i 


34-5 


33-5 


32.6 


31-7 


30- 7 


30.0 


293 




il 


26.7 


26.4 


26. 1 


25-8 


25.6 


25-3 


25.1 


27" 


I 


25-4 


25.1 


24.8 


24.6 


24-3 


24.1 


23-9 




3 
4 


26.4 


25-7 


25.0 


24.4 


23 -7 


23.2 


22.7 




2 

4 


29. 1 


27.9 


26.9 


25-9 


24.9 


24.1 


23-3 




1 
4 


34-0 


33-0 


32.1 


31.2 


30.2 


29-5 


28.8 




ii 


26.5 


26. 2 


25-9 


25.6 


25-4 


25.1 


24.9 


SO" 


4 
4 


25.2 


24.9 


24.6 


24.4 


24.1 


23-9 


23 -7 




3 

4 


26.2 


25-5 


24.8 


24. 2 


23 -5 


23.0 


22.5 




2 

4 


28.9 


27.7 


26.7 


25-7 


24.7 


239 


23.1 




1 
4 


33-5 


32.5 


31.6 


30-7 


29.7 


29.0 


28.3 




i| 


26.3 


26.0 


25-7 


25-4 


25.2 


24.9 


24.7 


36" 


4 

4 


25.0 


24.7 


24.4 


24. 2 


23-9 


23-7 


23-5 




3 

4 


26.0 


25 -3 


24.6 


24.0 


23-3 


22.8 


22.3 




I 


28.7 


27-5 


26.5 


25-5 


24-5 


23 -7 


22.9 




1 
4 


33-0 


32.0 


311 


30.2 


29. 2 


28.5 


27.8 




rl 


26.1 


258 


25 -5 


25.2 


25.0 


24.7 


24-5 



STEAM CONSUMPTION 



69 



TABLE 12 

Steam Consumption 

cross and tandem compound engines 





Load 


Pressure 


Stroke 


ioo-iio# 


I20-I30# 


I40-I50# 


IOO-IIO# 


I20-I30# 


i40-iso# 




Conde 


nsing 26" vacuum 


Noncondensing 


18* 


f 


15.6 


15-15 


14.8 


24.O 


23.0 


22.0 




3 

4 


16.75 


16.2 


15-5 


24-75 


22.7 


22.7 




3 
4 


17-5 


17-3 


17.2 


26.0 


24.0 


27.O 




i 


20.5 


20.1 


19.8 


41-5 


41.0 


40.5 




i£ 


16.2 


15-8 


15-5 


25.0 


24.0 


22.0 


21" 


I 


15-45 


150 


14.65 


23-5 


22.5 


21-5 




3 

4 


16.5 


16.O 


15-4 


24.2 


22.6 


21.6 




2 

4 


17-5 


17-4 


17-3 


26.4 


26.8 


27.1 




i 


22.o 


21 .O 


20.3 


41.2 


41 .O 


40.7 




if 


16. 1 


15-7 


14.9 


24-5 


23-4 


21.8 


24" 


f 


15-35 


I4.9 


14-55 


23.0 


22.0 


21.0 




3 

4 


16.2 


15-7 


15-2 


23-9 


22.6 


21.4 




2 


17.6 


17-5 


17.4 


26.8 


27.O 


27.2 




1 
4 


24.0 


22.5 


20.8 


40.9 


4O.9 


40.9 




I! 


16.0 


15-7 


15-3 


24.0 


23.O 


21.6 


27" 


4 

4 


151 


14-75 


I4.4 


22.5 


21-5 


20.5 




4 


16.0 


156 


I5-I 


23-3 


22.3 


21.3 




t 


17.7 


17.6 


17.6 


27.2 


27-3 


27-3 




1 
4 


26.0 


24.0 


21-5 


40.6 


40.9 


41. 1 




ll 


15-8 


15-5 


I5-I 


23-3 


22.3 


21.4 


3o" 


1 


150 


14.6 


14.25 


22.0 


21 .0 


20 ;o 




3 

4 


15-8 


15-4 


I4.9 


22.6 


21.8 


21. 1 




1 


17.7 


17.7 


17.6 


27.6 


27.4 


27.4 




1 
4 


28.0 


25.0 


22.0 


40.3 


40.8 


41-3 




l£ 


15-7 


15-3 


i5-o 


22.6 


21.8 


21.2 


36" 


4 
4 


14.6 


14.4 


14.0 


2i-5 


20.0 


19.0 




3 

4 


156 


152 


14.8 


22.0 


21-5 


21.0 




2 

¥ 


17.8 


17.8 


17.8 


28.0 


27-5 


27-5 




1 
4 


30.7 


26.6 


22.6 


40.0 


40.7 


41-5 




ll 


15 -5 


15-2 


14.8 


22.0 


21 . 2 


21 .0 



7o 



BASIC COSTS 



TABLE 13 

Steam Consumption 

horizontal single cylinder lentz engines 

Referred to i.h.p. at normal load and dry saturated steam 



Stroke 



Pressure 



85# ioo# ii5# I25# I4Q# i5o# i6o# 



Noncondensing 



18" 

14, 15, 16 and 17 x 21 
18, 19, 20 and 21 x 21 

24" 

27" 

30" 

33, 36, 42" 



18" 

14, 15, 16 and 17 x 21 
18, 19 and 21 x 21". . . 

24* 

27" 

30 

33, 36, 42" 



24.6 


23.6 


22. 7 


22. I 


21.3 


20.8 


20.3 


24.1 


23.I 


22. 2 


21 .6 


20.8 


20.3 


19 


8 


238 


22.8 


2I.9 


21.3 


20.5 


20.0 


19 


5 


234 


22.4 


21-5 


2O.9 


20. I 


19.6 


19 


1 


23.0 


22.0 


21 . I 


20.5 


19.7 


19.2 


18 


7 


22.6 


21.6 


20.7 


20. I 


19 3 


18.8 


18 


3 


22.2 


21 . 2 


2O.3 


19.7 


18.9 


18.4 


17 


9 



Condensing 



20 


3 


19 


7 


18 


7 


18 


2 


17-5 


17. 1 


16. 


19 


9 


19 


2 


18 


4 


17 


9 


17.2 


16.8 


16. 


19 


7 


18 


9 


18 


1 


17 


6 


16.9 


16.5 


16. 


19 


4 


18 


6 


17 


8 


17 


3 


16.6 


16.2 


15- 


19 


1 


18 


2 


17 


5 


17 





16.3 


15-9 


IS- 


18 


8 


18 





17 


2 


16 


7 


16.0 


15-6 


IS- 


18 


5 


17 


7 


16 


9 


16 


4 


15-7 


15-3 


14. 



For 50 degrees superheat reduce consumption 8%, 100 D superheat 12%, 
150 D 17%, 200 D 20%. 

At fractional loads, increase steam consumption as follows: 

LOAD il f f f i 

Saturated steam 5% 0% 4% 9% 25% 

Superheated steam 5% 0% 3% 7% 18% 



STEAM CONSUMPTION 



7i 



TABLE 14 

Steam Consumption 

horizontal cross and tanqem compound engines 

Referred to i.h.p., normal loads and dry saturated steam 





Pressure 


Stroke 


ioo# 


H5# 


I25# 


i 4 o# 


i5o# 


i6o# 


I75# 


ioo# 


200# 










Noncondensing 






„ 


18" 


22. I 


21 .2 


20.6 


19.8 


19-3 


18.8 


18. 1 


17-5 


17. I 


21" 


21.8 


2O.9 


20.3 


19-5 


19.O 


18.5 


i7-8 


17.2 


l6.8 


24" 


21-5 


20.6 


20.0 


19.2 


18.7 


18.2 


17-5 


16.9 


16.5 


27" 


21 .2 


2O.3 


19.7 


18.9 


18.4 


17.9 


17.2 


16.6 


16.2 


3o" 


20.9 


20.0 


19.4 


18.6 


18. 1 


176 


16.9 


16.3 


15-9 


33, 36, 42" 


20.6 


I9.7 


19. 1 


18.3 


18.8 


17-3 


16.6 


16.0 


i5-6 










Condensing 










18" 


16.4 


15-8 


1535 


14.8 


14-45 


14. 1 


136 


13.2 


12.9 


21" 


l6. I 


15-5 


15 05 


145 


14-15 


13-8 


13-3 


12.9 


12.6 


24' 


15.8 


15-2 


14-75 


14.2 


13-85 


13-5 


13.0 


12.6 


12.3 


27" 


15-5 


14-9 


14-45 


13-9 


13-55 


13.2 


12.7 


12.3 


12.0 


3o" 


15 -2 


I4.6 


14- 15 


13.6 


13-25 


12.9 


12.4 


12.0 


11. 7 


33, 36, 42* 


I4.9 


14-3 


13-85 


13-3 


12.95 


12.6 


12. 1 


11. 7 


11. 4 



For 50 degrees superheat reduce consumption 8%, for 100 D 12%, for 150 D 
17%, and for 200 D 20%. 

Increase steam consumption for fractional loads as follows: 



3 

4 


2 

4 


1 


3% 


7% 


20% 


2\% 


6% 


15% 



LOAD \\ f 

Saturated steam 6% 0% 

Superheated steam 6% 0% 



60. The Steam Consumption of simple low speed condensing 
engines is 18.5% greater than that of compound condensing 
engines. 

The Steam Consumption of simple low speed noncondensing 
engines is 42% greater than that of compound condensing 
engines. 



72 BASIC COSTS 

So also the Steam Consumption of simple high-speed non- 
condensing engines is 64% greater than that of compound 
-condensing engines. 

The foregoing steam consumption of compound condensing 
engines corresponds to the following thermal efficiencies, based 
on an initial normal water temperature of 52 F. 

TABLE is 

Thermal Efficiencies 

compound condensing engines 

I.h.p. 50 60 70 80 100 125 150 200 250 

Efficiency % 10.8 11. 1 11. 4 11. 6 12.0 12.4 12.6 13. 1 13.3 

Lkp 300 400 500 600 700 800 goo 1000 

Efficiency % 13.5 13.9 14.2 14.4 14.5 14.6 14.6 14.7 

This corresponds to the equation 

16.6 VM , x 

y ;vw^s'\ (25) 

where y = the efficiency in per cent, 
and M = the size in i.h.p. 

61. With superheated steam, the steam consumption is 
decreased. For 50 F. superheat, the steam consumption is 
reduced by 4%, for ioo° superheat 12%, for 150 superheat 
17 %, and for 200 superheat 20 %. This corresponds to increases 
in efficiency of 4% for 50 superheat, 8.4% for ioo°, 12.6% for 
150 , and an increase in efficiency of 16.6% for 200 F. super- 
heat, based on the efficiencies obtained for dry saturated steam. 
This shows that the efficiency increases in direct proportion to 
the increase in superheat, according to the equation: 

i = 0.0835 (26) 

where i = the increase in efficiency in per cent, 
s = the degrees superheat. 

The variation of consumption of dry saturated steam with 
variations of initial steam pressure is: 

(a) For compound condensing engines 

W = 41.6 — 0.074^, (27) 

where W = the pounds of steam used, 
and E = the temperature of the steam. 



FRACTIONAL LOADS 73 

(b) For low speed simple noncondensing engines 

W = 57.4 - 0.1E (28) 

The corresponding efficiencies are: 

(a) For compound engines 

y = o.o675(£ - 145) (29) 

(b) For simple low speed noncondensing engines we have 

approximately 

y = o.o466(£ — 120) (30) 

These formulae are obtained from the data above given. 
While, like all the above data, these efficiencies are based on 
factory guarantees, we do not consider them entirely satisfac- 
tory. They are, however, sufficiently accurate for ordinary 
practical use. 

62. Equipment is very rarely operated at full load. Nor- 
mally it is always run at some fraction of the full load. Far 
more important than full load efficiency are the efficiencies at 
fractional loads, for upon these, rather than those at full load, 
will the economy of operation depend. We therefore give the 
variation in steam consumption at fractional loads. These 
are as follows: 

(a) For compound condensing engines the per cent increase 
in steam consumption at various per cents of full load is as 
follows : 

TABLE 16 

Fraction of full load (F) 0.25 0.50 0.75 1.00 1.25 

Per cent full load, dry sat. steam (P S at.) 122 108 101 100 106 
Steam consumption, superheated steam 

GSsup.) 117 106 100 100 105 

The equations for these values are 

Ps "' = 57 .6 - S3HF - i.79)' (3I) 

and 



1 = 66.8 - 42.8FCF - 1.78) ' (32) 



sup. 



where (F) is the fraction (per cent) of full load. And the 
corresponding per cents of full efficiency are 

Vsat. = 57-6 - 53^ ~ i-79) (33) 

and y aup . = 66.8 - 42.SF(F - 1.78) (34) 



74 BASIC COSTS 

For low speed simple engines the values are: 

TABLE 17 

Fraction of full load (F) 0.25 0.50 0.75 1.00 1.25 

Per cent full load, dry sat. steam (i'sat.) 125 109 101 100 107 

Steam consumption, superheated steam (-Psup.) .. 117 106 100 100 105 



The equations for these values 

-Psat = — 



52 



P — 

■*■ sun. — 



6oF{F - i.8)' 

1 



(35) 
(36) 



66.8 - 42.SF(F - 1.78) * ' 
The corresponding per cents of full load efficiencies are: 
y sat = 52 - 6oF(F - 1.8) (37) 

and ;y SU p. = 66.8 - 42.8F(F - 1.78), (38) 

the latter being the same as for compound engines. 



35 



£25 

o 



20 



15 



E10 

B 
3 5 







































































































































































A.tte 


idanc 


e Co 


it pei 


-H.I 


.H. 




































































V 


^ 




















































^Ste 


amC 


onsu 


nptic 


n of 


Com 


30UIH 


Con 


d. Ei 


lgine 


5 





















































































































































































































































100 200 300 400 500 600 

I. H. P, 



700 800 900 1000 



Fig. 2. — Steam consumption and attendance costs of compound 
condensing steam engines. 

63. The attendance cost of steam engines alone is of com- 
paratively little value, as they must always be run as part of a 
system. Based, however, on the mean of the most authoritative 
data, this is about as follows: 



BOILERS 75 

TABLE 18 

I906-IO PRICES 

Size, i.h.p 10 20 30 40 50 60 75 100 

Attend., cts. per i.h.p. 0.47 0.38 0.30 0.275 °- 26 0.248 0.235 0-22 

Size, i.h.p 150 200 300 400 500 600 800 1000 

Attend., cts. per i.h.p. 0.192 0.172 0.152 0.137 0.127 °-i 2 0.108 0.100 

Size, i.h.p 1500 2000 

Attend., cts. per i.h.p 0.0885 0.08 

This corresponds approximately to the equation 

A= m (39) 

where A = attendance cost in cents per i.h.p. h., 
and M = the i.h.p. size of the engines. 

The cost of such incidentals as oil, waste, and supplies is 
about one-fourth of one per cent as great as the attendance cost. 

64. Boilers. — With present engineering development, there 
is no apparatus more used than the steam boiler. Under such 
conditions, it would be only natural to assume that the worth 
of the investment for such equipment would have been thor- 
oughly investigated. However, quite the opposite is the 
case. It is true that an enormous number of boiler tests have 
been made. But the results of these tests are so erratic that 
practically no confidence can be placed in them as far as the 
deduction of any general law of variation of efficiency with 
size is concerned. Measurements in heat are extremely diffi- 
cult, in fact so much so that the difficulty can hardly be exag- 
gerated. The same discrepancy appears in tests at fractional 
loads. 

Amongst the great mass of tests made, a few stand out as 
worthy of our entire confidence, though they are unfortunately 
too few to yield conclusive evidence. The most noteworthy 
of these is the test of the boilers of the Detroit Edison Com- 
pany, made by a corps of experts under D. S. Jacobus, the results 
of which are published in Vol. 33 of the American Society of 
Mechanical Engineers. The boilers had 23,650 sq. ft. of heat- 
ing surface each. They showed 79.3 % efficiency at rated load 



7 6 



BASIC COSTS 



and 80.5% at 70% of rated load. With increase of load, there 
was a steady fall in efficiency as far as the tests were conducted. 
Tests were also made of the same type of boiler (Stirling) of 
6400 sq. ft. of heating surface, by J. A. Hunter. They showed 
78% efficiency at rated load, 79.8% efficiency at 70% of rated 
load and 66% efficiency at 200% of rated load. These tests 
are well in consonance with the Jacobus tests. They are 
reported in full in Vol. 31 of the American Society of Mechanical 
Engineers. Further data of unusual value on boilers are given 



160 



50 



100 200 300 400 500 600 

Bo. H.P. 



700 800 900 1000 
Fig. 3. — Variation in boiler efficiency, with size. 



by A. D. Pratt in a paper that he presented before the Inter- 
national Engineering Congress at San Francisco, California, in 

1915- 

Taken as a whole, most manufacturers of boiler sell heating 
surface rather than boiler service. The results obtained from 
boilers depends in a very large measure on how the boiler is 
installed, the nature of the fuel used, and the methods of han- 
dling them. Good results can only be obtained where the 
entire boiler installation is made in accordance with the direc- 
tion of boiler experts. 

65. In the table below we give the factory prices of one make 
of boilers. These are at times somewhat erratic, although they 
are as a whole fairly well in proportion. To bring this out clearly 
we have added the last column of adjusted prices. Some of 
the difference between the factory and the adjusted prices is no 
doubt due to error in the cost calculations at the factory, but 
the greater part is due to commercial exigencies, such as the 



BOILERS 



77 



adaptation of the boiler dimensions to the use of stock sizes of 
sheets and tubes. The result of this is that the actual and 
rated sizes of boilers are not always the same. Allowing for 
this, the agreement between the factory and adjusted prices is 
excellent. 

TABLE 19 

Stationary Return Tubular Boilers With Full Front Setting 

1906-10 PRICES 







Cost factory, 




Adjusted cost 


Bo.h.p. 


Size 


total 


Per bo.h.p. 


per bo.h.p. 


IO 


30" x 7' 


$166.00 


$16.60 


$17-25 


12 


30" X 8' 


176.OO 


14 


•67 


I5-46 


IS 


36" X 8' 


207.00 


13 


80 


1366 


20 
20 


36" Xio' 
42" X 8' 


236.OO 

253.00 


II 
12 


80 1 

65/ 


11.87 


25 


36" xn' 


253.00 


IO 


12 1 


10.80 


25 


42" X io 7 


292.OO 


II 


68 J 




30 


42" X 12' 


313.OO 


IO 


43 


10.06 


30 


44" X 10' 


310.OO 


IO 


33) 




35 


44" X 12' 


340.OO 


9 


72 


9-57 


40 


44" X 14' 


375 -oo 


9 


37} 


9.17 


40 


48" X 12' 


397-5Q 


9 


94j 


45 


48" X 14' 


435- 00 


9 


67 


8.90 


50 


48" x 1 6' 


472.00 


9 


44l 


8.65 


50 


54" X 12' 


465.00 


9 


30/ 


60 


54" X 14' 


508 . 00 


8 


47 I 


8.29 


60 


60" XI2' 


550.00 


9 


I 7J 


65 


54" X 15' 


530.00 


8 


18 


8-i5 


70 


54" x 16' 


555 00 


7 


93 1 


8.04 


70 


60" X 14' 


600.00 


8 


57l 


75 


60" x 15' 


622.00 


8 


33 


7-94 


80 


60" x 16' 


650.00 


8 


15 


7.84 


85 


66" x 14' 


720.00 


8 


50 


7.76 


90 


66" x 15' 


730.00 


8 


11 


7.70 


100 


66" x 16' 


770.00 


7 


70 


7-57 


no 


66" x 18' 


820.00 


7 


45 


7.48 


125 


72" X 16' 


920 . 00 


7 


36 


7-36 


150 


72" X 18' 


1000 . 00 


6 


67 


7.22 



The equation for the adjusted prices is 



C = 



M 



+ 6.50, 



where C = cost in dollars per bo.h.p. 
and M = the size in bo.h.p. 



(40) 



78 



BASIC COSTS 



TABLE 20 

Water-tube Boilers with Full Front Setting 

prices, 1906-10 



Size, 


bo.h.p 


100 


125 


150 


175 


200 


250 


Price 


1 per bo.h.p. . . 


• • $1360 


$12.70 


$12.10 


$11.67 


$ii.35 


$10.90 


Size, 


bo.h.p 


300 


35o 


400 


4SO 


500 


600 


Pria 


; per bo.h.p. . . 


. . . $10.60 


$10.39 


$10.22 


$10.10 


$10.00 


$9-85 



The equation for these values is 



45o 
M 



+ 9.10. 



(41) 



Fernold and Orrok, " Engineering of Power Plants," give 

c =f+' 



(42; 



and the cost of setting as 

C s = 140 -f 2M (43) 

66. Cost of Building. — These same authorities, Fernold and 
Orrok, give average data on the cost of buildings, and the divi- 



$20 










































20 

C-l 


















































































X 
6 15 

PQ 

Im 


\ 
































































W 


ter : 


'ube 


Boile 










lio 




























rs 






















Ret 


urn ' 


^ibul 


»r Be 


ilers 


















5 





























































































































25 50 75 100 125 150 175 200 225 250 

Bo. H.P. 

Fig. 4. — Variation in prices of boiler with size and type. 

sion of costs in power plants in good practical form, some of 
which is given below. 



BUILDINGS 79 

TABLE 21 
Cost of Buildings, 1906-10 

Type Cost in dollars Cost in cents 

per sq. ft. per cu. ft. 

floor surface of contents 

Mill construction o. 80 to 1 . 10 6. 5 to 8.5 

Fireproof stores, factories and warehouses with 

brick, concrete, stone and steel construction. . . 2.00 to 3.00 14.0 to 25.0 
Concrete and reintorced concrete shops, factories 

and warehouses 1 . 25 to 1 . 75 8.0 to 16.0 

Plain power houses, with concrete floors, brick and 

steel super-structure 2 . 00 to 2 . 75 9 . o to 12 . o 

Power houses under city conditions with superior 

architecture 3 .00 to 4. 50 15 .0 to 30.0 

Exact data cannot possibly be given, because the cost of 
buildings varies enormously with local conditions. These must 
always be determined by the engineer before reaching any final 
decision. 

TABLE 22 

Cost of Steam Power Plant Buildings 

1906-10 

PER ENGINE HORSE POWER 

Simple Noncondensing Engines 

H.p 10 12 15 20 30 40 50 75 

Boiler house $37- *5 $33 00 $28.50 $24.50 $20.50 $18.00 $16.00 $13.00 

Engine house 4.80 4.35 3.90 3.30 2.75 2.50 2.30 2.15 

Coal bins 20.00 18.00 16.00 13.70 11.00 9.80 8.30 6.00 

Simple Condensing Engines 

H.p 10 12 15 20 30 40 50 75 100 

Boiler house $33.70 $29.60 $26.20 $21.60 $18.20 $16.00 $14.80 $11.30 $9.70 

Engine house. .. . 14.40 12.60 10.90 8.60 7.75 6.40 5.35 4.90 4.30 
Coal bins. ..... . 19.00 17.90 15.80 13.60 11.00 8.70 8.50 6.30 5.70 

Compounding Condensing Engines 

H.p 100 200 300 400 500 600 700 800 

Boiler house 1ft « (11.20 8.00 6.40 5.70 5.35 5.00 

Engine house J ' 5 ° 24 ' \ 11.20 9.35 8.50 7.20 6.30 5.60 

Coal bins 5.70 4.00 3.10 2.60 2.40 2.25 2.10 2.05 

H.p 900 1000 1500 2000 

Boiler house $4.70 $4.55 $410 $3-95 

Enginehouse 5.35 500 4.75 4.55 

Coal bins 1.95 1.80 1.75 1.66 



80 , BASIC COSTS 

The total cost of engine and boiler house and coal bins for 
simple noncondensing engines is given by the equation 

G = 63.5 - § V200M - M 2 - 1900. . . (44) 
For simple condensing, this is 



C 2 = 65.0 — I V200M — M 2 — 1900, . . (45) 
while for compound condensing engines the approximate 
equation is 

Ca=^ + 8. 2 (46) 

67. The cost of complete steam installations is given as 
follows : 

TABLE 23 

H.p • 5000 10,000 20,000 30,000 40,000 50,000 

Cost per h.p $120.00 $100.00 $90.00 $80.00 $75.00 $70.00 

These costs are found to be divided about as follows: 

TABLE 24 

Percentage of total 

Buildings, real estate and excavations 14.6 

Turbines and generators 23 . 5 

Condensers 5.7 

Boilers, superheaters, stokers and stacks 28. 7 

Bunkers and conveyors 4.8 

Boiler, feed and service pumps 1 . o 

Feed water heaters 1.6 

Switchboard and wiring 3.5 

Excitors 2.1 

Foundations — machinery 1 . 1 

Piping and conduit 7.1 

Crane 1.5 

Superintending and engineering 4.8 

68. Centrifugal Pumps. — The variation in price of one 
certain make of centrifugal pumps with size and type is given 
below. 



BELTED CENTRIFUGAL PUMPS 



81 



TABLE 25 

Low Head Horizontal Belted Centrifugal Pumps 
1910-12 



Size 
inches 


Normal capacity, 
g.p.m. 


Cost of pump 


Cost per Adj 
1000 g.p.m. cc 


jsted 
sts 


I 


20 


$27.00 


$1350.00 




I* 


SO 


33- 00 


660.OO 






2 


100 


42.00 


420.OO 






,1 
2 2 


ISO 


51.00 


340.OO 






3 


225 


60.00 


266.OO 






3h 


300 


69.00 


230.OO 






4 


400 


75.00 


187.50 






5 


700 


100.00 


142 . 86 






6 


900 


120.00 


133-33 






7 


1200 


150.00 


125.00 






8 


1600 


180.00 


112.50 $18 


3. 70 


10 


3000 


240 . 00 


80.00 25 


3.IO 


12 


45oo 


300 . 00 


66.67 33 


3-75 


14 


6000 


425.00 


70.83 40 


3.20 


iS 


7000 


475 00 


67.86 45< 


5.90 


16 


8000 


525.00 


65.62 51 


r .60 


18 


10,000 


900.00 


90.00 61, 


5.00 


20 


14,000 


1000 . 00 


71.40 82 


[.80 


24 


20,000 


1200.00 


60.00 113 


2.00 


30 


30,000 


1550.00 


51.67 164* 


).oo 


36 


40,000 


2400 . 00 


60.00 2l6< 


3.00 


40 


50,000 


2700.00 


54 . 00 268 v 


5.00 


44 


60,000 


3200.00 


53-33 320c 


3. OO 



The factory costs are somewhat erratic due to the lack of 
knowledge of exact costs on the part of the manufacturer. 
The prices on the larger sizes are no doubt based on " esti- 
mates." The equation based on the adjusted values, i.e., those 
that have been brought into consonance with themselves, is 



where C = cost in dollars per 1000 g.p.m. of capacity, 
and M = the capacity in iooo's of g.p.m. 



(47) 



82 



BASIC COSTS 



bwu 










































600 




1 














































































500 


















































































400 




\ 














































































300 




y 






« 


































\ 






































200 




\ 










































\ 










































\ 


s 








'Hig 


h He 


ad B 


sited 


Primps 






































Low Head Belted 


Pumps v 





















































12 16 20 24 28 32 36 40 

Size of Pump in Inches of Discharge Nozzle 

Fig. 5. — Variation in price of centrifugal pumps, with size and type. 



TABLE 26 

High Head Horizontal Belted Centrifugal Pumps 
iQio-12 



Size 


Normal capacity, 


Cost, factory 


Cost per 


Adjusted 


inches 


g.p.m. 


1000 g.p.m. 


costs 


if 


SO 


$67 . 20 


$1344.00 


$77.625 


2 


IOO 


84.00 


840.00 


83-75 


2\ 


I50 


94.80 


632.00 


90.375 


3 


225 


112.20 


500 . 00 


99-94 


4 


400 


I34-40 


338.OO 


122.25 


5 


700 


150.00 


214.30 


160.50 


6 


900 


198 . 00 


220.00 


186.00 


7 


I200 


240 . 00 


200 . 00 


224.25 


8 


1600 


294 . 00 


183.75 


275-25 


10 


2500 


390.00 


156.00 


390.00 


12 


4000 


522.00 


130.50 


581.25 



The equation based on the adjusted values is 
C=^ + $127.50. . 



(48) 



DIRECT-CONNECTED CENTRIFUGAL PUMPS 



83 



TABLE 27 

Direct-connected Horizontal Centrifugal Pumps Less Motor 
1910-12 

25 Ft. Head 



H.p. 


R.p.m. 


Size 


Capacity 


Cost 


Cost per 


inches 


g.p.m. 


factory 


1000 g.p.m. 


1 
2 


1800 


3 

4 


15 


$70 . OO 


$4666.67 


I 


1800 


3 


70 


IOO.OO 


1428.OO 


2 


1800 


3§ 


130 


125.OO 


960.OO 


3 


1800 


4 


240 


140.OO 


580.00 


5 


1800 


5 


420 


170.OO 


405.OO 


7§ 


I200 


5 


6lO 


230 . OO 


376.OO 


7l 


I200 


6 


650 


310.OO 


476.OO 


10 


I200 


5 


820 


240.OO 


294 . OO 


10 


I200 


8 


IOOO 


385 • °o 


385-00 


15 


900 


7 


1300 


3IO.OO 


237.00 


20 


1800 


7 


1350 


33O.OO 


245.00 


20 


720 


10 


2300 


500.00 


218.00 


25 


720 


10 


2500 


530.00 


212.00 


.30 


1800 


8 


2000 


385-00 


192.00 


30 


900 


10 


•320O 


500 . OO 


156.00 


35 


600 


12 


3500 


630 . OO 


180.00 


50 


900 


12 


4800 


675.00 


144.00 


50 


600 


12 


480O 


750.00 


156.00 



8 4 



BASIC COSTS 



TABLE 28 

Direct- connected Horizontal Centrifugal Pumps Without Motor 

1910-12 

50 Ft. Head 



H.p. 


R.p.m. 


Size 
inches 


G.p.m. 


Cost 


Cost per 
1000 g.p.m. 


2 


1800 


I 


25 


$80.00 


$3200.00 


2 


1800 


2 


50 


100.00 


2000.00 


3 


1800 


2 2 


IOO 


115.00 


1150.00 


5 


3600 


3 


200 


I35-QO 


675.00 


5 


1800 


2| 


200 


145.00 


725.00 


1\ 


1800 


4 


350 


180.00 


512.00 


10 


1800 


4 


500 


185.00 


370.00 


IS 


1800 


4 


700 


210.00 


300.00 


20 


I200 


6 


IOOO 


270.00 


270.00 


20 


1800 


7 


I IOO 


330 . 00 


300.00 


25 


900 


7 


1250 


400.00 


320.OO 


30 


1800 


8 


1500 


385 • 00 


260.00 


35 


900 


8 


2000 


420 . 00 


210.00 


40 


900 


• 8 


2200 


420.00 


191.00 


40 


900 


12 


2600 


930.00 


357-OQ 


5o 


900 


10 


2500 


530 . 00 


212.00 


75 


900 


10 


3900 


550.00 


141.00 


100 


900 


12 


5500 


650 . 00 


118.00 


125 


600 


15 


7000 


1000 . 00 


143.00 


150 


720 


14 


7750 


900 . 00 


116. CO 



DIRECT-CONNECTED CENTRIFUGAL PUMPS 85 



TABLE 29 

Direct-connected Horizontal Centrifugal Pumps Without Motor 

1910-12 

100 Ft. Head 



H.p. 


R.p.m. 


Size 
inches 


G.p.m. 


Cost 


Cost per 
1000 g.p.m. 


3 


3600 


I 


43 


$140.00 


$3260.00 


5 


3600 


3 


135 


I35-QO 


IOOO.OO 


72 


1800 


2 


125 


150.OO 


I 200 . OO 


15 


3600 


3* 


300 


150.00 


500 . OO 


IS 


1800 


2\ 


315 


185.00 


588.OO 


20 


1800 


4 


470 


210.OO 


446.OO 


25 


1800 


4 


560 


235.00 


420 . OO 


30 


1800 


5 


700 


260.OO 


37I.OO 


35 


1800 


4 


700 


240 . OO 


342.00 


40 


I200 


5 


1040 


325.OO 


313.OO 


50 


1800 


6 


1275 


270.OO 


2I2.00 


75 


I200 


8 


1800 


420.OO 


233.OO 


75 


1200 


7 


2000 


530.OO 


265.OO 


100 


1200 


8 


2450 


550.00 


217.OO 


100 


900 


8 


2700 


640 . OO 


237.OO 


150 


1800 


12 


3800 


IOOO . OO 


264.OO 


150 


900 


10 


4000 


680 . OO 


i 70 . OO 


200 


I200 


12 


5600 


680 . OO 


121.00 


250 


900 


14 


7000 


1100.00 


157.00 



$3500 




100 125 150 
H.P. of Pump 



Fig. 6. — Prices of direct-connected centrifugal pumps. 



86 



BASIC COSTS 



TABLE 30 

Direct-connected Horizontal Centrifugal Pumps Without Motor 

1910-12 

150 Ft. Head 



H.p. 


R.p.m. 


Size, 
inches 


G.p.m. 


dost 


Cost per 
1000 g.p.m. cap. 


3 


1800 


I 


17 


$230.00 


$13,530.00 


5 


1800 


1 


30 


230.00 


7,670.00 


Ih 


1800 


li 


50 


270.00 


5 ,400 . OO 


10 


1800 


2 


90 


270.00 


3,000.00 


15 


3600 


3* 


150 


150.OO 


I,000.00 


20 


1800 


2\ 


160 


190.OO 


1,190.00 


20 


1800 


4 


300 


430.00 


1,430.00 


25 


1800 


3 


235 


200.00 


850.OO 


35 


1800 


4 


500 


450.OO 


900.OO 


35 


1800 


5 


550 


575 00 


1 ,042 . OO 


40 


I20O 


4 


430 


415.00 


965.00 


50 


1800 


5 


800 


600.00 


750.OO 


50 


1800 


6 


850 


725.00 


850.OO 


50 


I200 


5 


600 


510.00 


850.06 


75 


I200 


6 


1200 


500.00 


416.OO 


75 


1800 


8 


1400 


900.00 


643.00 


100 


I200 


8 


1500 


550-00 


366.OO 


100 


1800 


8 


1800 


925.00 


51300 



VERTICAL CENTRIFUGAL PUMPS 



87 



$900 



800 



700 

































































































































































/' 






























e»<* 


Py/ 






































♦a 




©? -^ 






































u^' 










































^ 




«-«€ 


»x 












































































































































































































































































: 


L 


a 


y 




\ 


i 


l 










7 


i 


1 




d 


10 



a 

a 
«6oo 

a 
I 
^500 



400 



300 



200 



Size of Punrp in Inches of Discharge Nozzle_Diameter 
Fig. 7. — Prices of vertical centrifugal pumps, complete. 



TABLE 31 

Vertical Centrifugal Pumps 
1910-12 

Complete with frames, valves, shafting, boxes, pipe, collars, etc. 



Size, 


Normal 








Inches 


Capacity 












With 25 ft. frame 


With 50 ft. frame 


With 75 ft. frame 


2 


100 g.p.m. 


$196.00 


$265.00 


$334 OO 


2\ 


150 " 


22I.OO 


293 . 00 


364.OO 


3 


225 " 


249 . OO 


332.00 


408.OO 


3^ 


300 " 


281 . OO 


368 . 00 


452.OO 


4 


400 " 


300 . OO 


397.00 


482.OO 


5 


700 " 


353 00 


462 . 00 


561.OO 


6 


900 " 


423.00 


550-°° 


658.OO 


7 


.1200 " 


496 . OO 


616.00 


733.00 


8 


1600 " 


605 . OO 


740.00 


860 . OO 



88 BASIC COSTS 

69. Cost of Large Pumping Units. — The following quotations 
were made on various types of complete pumping plant f.o.b. factory, 
between 1908 and 1910. 

1. A 25,000 g.p.m. horizontal pump for a total head of 16 ft., direct- 
connected to a 150 h.p. 257 r.p.m. a.c. motor together with primer, etc., 
and 100 ft. of piping. Price quoted $6000. 

2. Same as (1) above but for 20 ft. head. Price quoted $7500. 

3. A 25,000 g.p.m. horizontal pump for a total head of 16 ft., direct 
connected to a 150 h.p. horizontal tandem compound condensing steam 
engine, complete with water tube boiler, jet condenser, and necessary 
piping. Price quoted $12,000. 

4. Same as (3) but for 20 ft. head. Price quoted $15,000. 

5. One 50,000 g.p.m. horizontal, double suction pump, direct-connected 
to one 300 h.p. 200 r.p.m. a.c. motor for 16 ft. total head, together with 
100 ft. of piping and electric primer. Price quoted $9000. 

6. Vertical axial flow turbine with a capacity of 125 cu. sec. ft. against a 
head of 6 ft., with 69 in. runner, speed 65 r.p.m. direct geared to a 150 h.p. 
485 r.p.m. vertical motor and having a guaranteed efficiency of 60%. 
Price quoted $6100. 

7. Same as (6) but for 75 cu. sec. ft. of capacity and 57 in. runner and 
100 h.p. motor. Price quoted $4900. 

8. One 40 in. horizontal centrifugal pump having a capacity of 125 cu. 
sec. ft. against a total head of n ft., belted to a 250 h.p. motor. Pump 
efficiency guaranteed 50%. Price quoted $6100. 

9. Same as (8) but for 6 ft. head, 50% efficiency, and belted to a 150 
h.p. motor. Price quoted $3400. 

70. Operating Costs of Centrifugal Pumps. — The material 
consumed in the case of pumps is power together with a very 
little oil and waste. The amount of power used depends upon 
the pump efficiency and the cost of the power or its price. The 
efficiency of pumps varies both with the size of the pump and 
the head pumped against, larger pumps and greater heads 
giving higher efficiency, according to the following table. 



OPERATING COSTS OF CENTRIFUGAL PUMPS 89 



TABLE 32 

Efficiencies 

centrifugal pumps 







Head in Feet 




Capacity, 








g.p.m. 


















io' 


12' 


15' 


20' 


25' 


30' 


40' to 70' 




% 


% 


% 


% 


% 


% 


% 


50 


22 


23 


24 


25 


29 


30 


33 


75 


25 


26 


27 


29 


33 


35 


36 


100 


28 


30 


32 


34 


36 


38 


40 


150 


32 


34 


35 


37 


40 


42 


45 


200 


36 


37 


39 


42 


45 


48 


50 


300 


40 


4i 


43 


45 


47 


5i 


53 


400 


42 


44 


46 


48 


50 


54 


56 


500 


44 


46 


48 


50 


53 


56 


58 


600 


46 


48 


5o 


52 


54 


58 


60 


700 


48 


50 


52 


54 


57 


60 


62 


800 


50 


52 


54 


56 


61 


63 


64 


900 


52 


54 


56 


59 


62 


63 


64 


1,000 


53 


55 


57 


59 


62 


64 


65 


1,200 


54 


56 


58 


60 


64 


65 


66 


1,400 


55 


57 


59 


62 


65 


67 


68 


1,600 


57 


59 


61 


63 


65 


67 


68 




Pump Efficiency Continued 




5' 


IO' 


15' 


25' 


35' 


50' 




% 


% 


% 


% 


% 


% 


6,000 


50 


60 


65 


70 


70 


70 


7,000 


50 


61 


65 


70 


70 


70 


8,000 


50 


62 


66 


70 


70 


70 


10,000 


50 


63 


67 


70 


73 




15,000 


5° 


63-5 


68 


70 


74 




20,000 


50 


64 


69 


70 


75 




30,000 to 














60,000 


50 


65 


70 


70 







go 



BASIC COSTS 



70 



50 



§40 



30 



20 



10 

















































To" 


ft. B 


ead 
















































10 f 


t. He 


ad 

















































































































































































































































































































































































































































































































2000 4000 6000 8000 10000 12000 14000 16000 18000 20000 
G. P. M. 

Fig. 8. — Efficiencies of centrifugal pumps. 
At io ft. head, the variation of pump efficiency with capacity 



is 



67- 



14 

VM" 



(49) 



where y = efficiency in per cent, 

and M = the pump capacity in thousands of gallons per 
minute capacity. 
With increased head the pump efficiency increases by the 
following per cents: 

Head 10 12 15 20 25 30 40 50 

Per cent increase in efficiency. .. . o 4 9 15 18.5 21 25 28 



Corresponding to the equation 



160 



(So) 



where y p = per cent increase in efficiency, 
and h = the head in feet. 

The lubricating oil used amounts to about o.ooi pint per 
water horse power hour. 



FRACTIONAL LOAD EFFICIENCY 91 

71. It is often assumed that the cost of attendance of cen- 
trifugal pumps is zero. This is hardly correct, although this 
cost for centrifugal pumps is very small. It depends upon the 
general design of the installation normally being about 

*-2SL (51) 

where A = cost of attendance in cents per h.p.h. as of 1910, 
and M = the h.p. required to drive the pump. 

It must be borne in mind that all costs are as of 1906-10. 
To get the attendance cost as of today divide by the relative 
value of gold now as compared with the date above, as indi- 
cated by the relative purchasing power. Thus if the purchasing 
power of gold now is only half of what it was then, all costs 
now would be double that given above. 

Based on the study of the efficiency of pumps at fractional 
loads covering the tests of several hundred pumps varying in 
size from 1 inch to 12 inches and consisting of four different 
makes and some twelve different types, we have obtained the 
following average values of the per cent of full load efficiency 
at fractional loads viz. : 

Per cent of full load o 25 50 75 100 125 150 

Per cent of full load efficiency. . . o 43.6 75 93.7 100 93.7 75 

Corresponding to the equation 

y p = F(2 - 0.01F) (52) 

where y p = per cent of full load efficiency, 
and F = per cent of full load efficiency, 



92 BASIC COSTS 

72. Motors and Generators. 

TABLE 33 

Induction Motors with Starters 

prices, 191 2 

2 and 3 phase 60 cycles 110-220 volts, squirrel cage type 



Size, 
H.p. 



2 

I 
2 

3 
5 

l\ 
10 

15 

20 

25 
30 
40 

50 

75 

100 

150 



1800 



$33 00 

40.00 

50.00 

57.00 

70.00 

135 00 

160.00 

192.00 

235.00 

255-00 

305.00 

350.00 

400 . 00 

51500 



R.P.M. 



$40 . 00 

50.00 

58.00 

71 .00 

90.00 

160.00 

195.00 

225.00 

272.00 

305.00 

340.00 

390 . 00 

440 . 00 

565 00 

680 . 00 



$47.00 
63.00 
75.00 
94.00 
1 1 7 . 00 
195.00 
230.00 
270.00 
305.00 
340.00 
375- 00 
435- 00 
485.00 
610.00 
724.00 
960.00 



720 


600 


$77.00 

91 .00 




$110.00 


115.00 


140.00 


143.00 


1 70 . 00 


220.00 


267 . 00 


265 . 00 


300 . 00 


306 . 00 


348.00 


330 . 00 


390.00 


377.00 


430.00 


420 . 00 


465 . 00 


481 .00 


535- 00 


535 00 


595- 00 


664 . 00 


720.00 


780 . 00 


832.00 


1010.00 


1070.00 

















































$1200 


















































































1000 


















































































Motors 

GO 

O 
O 






















































ftf 


























Price of 

O 
















<£? 


J^ 
































































400 








s 


/ 


f^\ 


1> 






















, 










/ 


/ 


s$ 


^ 






























200 


j 


/, 


/ 




































Li 


/ 









































T 









































20 40 60 80 100 120 140 

H. P. 

Fig. 9. — Prices of squirrel cage motors. 



160 



180 200 



MOTORS 



93 



73. 



TABLE 34 

Induction Motors with Starters 

prices, 1910-12 

Slip-ring type 110-550 volts, 3 phase 60 cycle 



Size, 


R.P.M. 


H.p. 


1800 


1200 


900 


720 


600 


1 


$60.00 

80.00 

104.00 

122.00 

155-00 

210.00 

235.00 
275.00 


$77- 00 
104.00 
113.00 
138.00 
175.00 
224.00 
260.00 
300 . 00 
340.00 
380 . 00 
425.00 
505-00 
575 00 
725.00 










$113.00 
136.00 
164.00 
188.00 
250.00 
280.00 
330.00 
380 . 00 
425.00 
475- 00 
562 . 00 
640 . 00 
790 . 00 
920.00 












3 
5 
1\ 














$300.00 
360.00 
410.00 
468.00 
S1500 
608 . 00 
680.00 
830 . 00 
960.00 




15 

20 

25 
30 
- 40 
50 
75 
100 


$380.00 
440.00 
495.00 
550.00 
640 . 00 
710.00 
860 . 00 








990 . 00 









As previously noted, these prices are as of 19 10-12. To 
obtain, for example, prices as of 1917, divide by 0.65. 



94 



BASIC COSTS 



74. 



TABLE 35 

Direct Current Motors 

prices, 1910-12 

Open shunt wound type 115-230 volts 



H.p. 


Speed 


Price 


1 


1500 


$65.00 




i-5 


1400 


75.OO 




2-5 


1400 


IOO.OO 




3-5 


1700 


IOO.OO 




3-5 


1300 


128.00 




5 


I550 


128.OO 




5 


I IOO 


IOO.OO 




7-5 


1500 


160.OO 




7-5 


IOOO 


200.00 




10 


1400 


205 . OO 




10 


I IOO 


233.00 




15 


1400 


245.00 




15 


IOOO 


284 . OO 




20 


1350 


284.OO 




20 


900 


346.00 




25 


I200 


346.00 




25 


800 


440.00 




30 


IOOO 


440.OO 




30 


700 


500.OO 




35 


950 


516.OO 




35 


700 


570.OO 




40 


950 


500 . OO 




40 


600 


665 . OO 




50 


IOOO 


600.00 




50 


600 


785 . OO 





Note. — The price of compound wound motors is 5% higher. 



75. 



MOTORS AND GENERATORS 



95 











































$1000 


















































































800 

u 

1 

O 

2 600 
o 

4-> 


























































«& 






































V\ 




s*b 


y 


































^ 


& 


\ry 






















o 

o 
400 


















































































200 




y 


t 


/- 


































A 


'/ 






































n 


? 









































10 20 30 40 50 60 70 80 90 100 

- H. P. 

Fig. io. — Prices of direct current motors. 













































$30 


















































































25 


















































































20 


















































































> 

«15 

u 


















































































10 


















































































5 






























































































































100 200 300 400 500 600 700 800 900 1000 
K. V. A. 

Fig. ii. — Prices of A. C. generators. 

TABLE 36 

A. C. Generators 

prices, IQIO-I2 

60 cycle three-phase 220-550 volts. Direct-connected type 

Size, kv.a 10 25 50 75 100 150 250 500 750 1000 

Price per kv.a . . .$25.00 20.00 16.50 14.75 14.00 12.85 II -5° 10.00 9.60 9.40 



96 BASIC COSTS 

This corresponds almost exactly to the equation 

c =^ + - 

where C = cost in dollars per kv.a., 
and M = the size in kv.a. 



(53) 



TABLE 37 

Transformers 

prices, 1910-12 

1 100-2200 to 110-220 volts 

1 i-5 2 2.5 3 

Price $26.00 32.00 39.00 46.00 51.00 57.00 

Size, kw.. 10 15 20 25 30 

Price $13300 184.00 229.00 269.00 314.00 



Size, kw. . . .0.6 



4 
68.00 



5 

80.00 

40 
395 00 



7-5 
108 . 00 

50 
474.00 



76. Operating Costs of Electric Machinery. — The efficiency 
at full lead of three phase, squirrel cage motors, no to 550 
volts, is independent of the speed but varies with the size as 
given below. 

































































































































$500 


















































































400 
2 


















































































<2 300 

BO 

c 


















































































H 
^200 

■s 

a 
100 














































































































































































































10 15 20 25 30 35 

Kw. 

Fig. 12. — Prices of transformers. 



40 



45 



50 



TABLE 38 
Efficiencies 
Induction motors 3 phase 60 cycle 

Size in h.p 1 2 3 5 7.5 10 15 20 25 

Efficiency in per cent.. .78.4 82.5 84.2 85.8 87.2 87.9 88.6 89.1 89.4 

Size in h.p 30 35 40 50 60 75 100 150 200 250 

Efficiency in per cent.. .89.7 89.85 90.05 90.25 90.4 90.6 90.8 91. 1 91.24 91.33 



OPERATING COSTS OF ELECTRIC MACHINERY 97 



150 
140 
130 
120 
110 
100 



0% 
I 80 



10 

































































































































































































































































Efl 


• = t 


2.2- 


13.8 

VhT 


i 
























































• 












































































*/* 




































J lift* 


































?.i : . ,/ : 


































'< x. •#< 
























































J 










- — 




* 


* 













72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 8* 
Per cent. Efficiency 

Fig. 13. — Efficiencies of A-c. 60 cycle motors. 



89 90 91 



This corresponds to the equation 



92.2 — 



13-8 



(54) 



where y = the per cent efficiency, 
and M = the size in h.p. 

It should be borne in mind that such data as we present are 
based on the study of the results of tests of our very best 
and largest manufacturers. The data themselves, together 
with the curve of values adopted, is shown in Fig. 13. The 
discrepancies in the data are due to redesign of some units be- 
fore the like service could be given to others; and to some 
error in test, which, though small, is nevertheless quite signifi- 
cant and finally to commercial exigencies such as the use of a 
frame designed for one given size and speed, for a different 
size and speed. Such adaptations of one frame to several sizes 
generally give greater economy of production but at a re- 
duction in worth of the unit due to reduced efficiency and 
power factor. 



9 8 



BASIC COSTS 



The efficiency of generators at full load for 3 phase alter- 
nating current of 60 cycles at from 200 to 550 volts varies 
with size as follows: 



TABLE 39 

Efficiencies of A-c. Generators 

60 cycles 220 to 550 volts 

Size, kv.a 50 75 100 125 150 200 300 500 

Per cent efficiency 88.5 90.0 90.9 91.5 91.9 92.6 93.2 93.9 

Size, kv.a 750 1000 1250 1500 2000 3000 4000 5000 

Per cent efficiency 94.4 94.7 94.9 95.0 95.2 95.4 95.5 95.6 



These values correspond to the equation 

55 



96.4 - 



VM 



(55) 



77. At fraction loads the per cent of full load efficiency 
obtained from motors and generators is as follows: 



100 



97 

96 

1 95 

a» 
|94 

Ed 

^'93 

a 

*91 
90 

89 
88 
87 

86, 






500 1000 1500 2000 2500 3000 3500 4000 4500 5000 
K. V. A. 



Fig. 14. — Efficiencies of a-c. generators. 



INTERNAL COMBUSTION MOTORS 99 

TABLE 40 

Per Cent of Full Load Efficiency of Induction Motors 

3 phase 60 cycle 210-550 volts 

Per cent of full load 125 100 75 50 25 

Per cent of full load efficiency 97 100 100 97 90 

These values correspond to the equation 

y , = I0O - fc^- 2 ( S 6) 

400 
where y p = per cent of full load efficiency, 
and F = per cent of full load efficiency. 

TABLE 41 

Per Cent of Full Load Efficiency of A-c. Generators 

3 phase 60 cycle 220-550 volts 
Per cent of full 

load 125 ioo 90 80 70 60 50 40 30 25 

Per cent of full 

load efficiency 98 . 5 100 99.5 98.5 97.0 95.0 92.0 89.0 85.0 82.5 

This corresponds to the equation 

(100 - F) 2 , x 

y p = 100 - ^ '-. .... (57) 

320 

The attendance cost of motors and generators of 19 10-12 is 
approximately 

A = 2^5 ( 8) 

VM 

78. Internal Combustion Motors. — The prices of gas 
engines depends on whether they are two or four cycle, Otto 
cycle, Diesel, or semi-Diesel, as well as on the design and 
workmanship. Some engines are needlessly elaborate, others 
so unsubstantially built that they are made to sell rather than 
to use. The two stroke engine, except for the Diesel or semi- 
Diesel cycle, is not of practical value, and no data thereon is 
therefore given. 

The sizes and corresponding prices (191 2) for one make of 
well-designed and well-built type of gas engines is given below: 



IOO 



BASIC COSTS 



TABLE 42 

Stationary Oil Engines 

prices, 1910-12 
Heavy Duty — Otto cycle 





Horizontal 


Vertical 




No. cy. 


Total 


Per h.p. 


No. cy. 


Total 


Per h.p. 


2 
3 
4 
6 
8 


1 
1 
1 

1 

T 

I 
I 
I 


$180.00 
270.OO 
360 . OO 
500 . OO 
630 . OO 
810.OO 
I080.OO 
1350.OO 


$90 . OO 
90.OO 
90.OO 
83 -33 
78.75 
67.50 
60.OO 
54.OO 


1 


$180.00 


$90 . OO 


1 
1 


270.OO 
360 . OO 


67 . 50 
60.OO 


12 
18 


1 


720.00 


60.OO 


25 

25 
32 
36 
40 

50 


1 
2 


1080.00 
1350.00 


43-25 
54.00 


I 

2 

I 

i or 2 


1660.OO 
2160.OO 
1980.OO 
2520.OO 


52.OO 
60.OO 

49 50 
50.40 


3 


1980 . 00 


55- 00 


2 

1 


2160.00 
2520.00 


43.20 
50.40 


64 

75 
75 
80 

IOO 


2 


3240.OO 


5°-50 


3 

1 
1 or 2 


3260.00 
3870.00 
5000 . 00 


43 SO 

51.60 
50.00 








2 
2 


3870.OO 
5000 . OO 


48.38 
50.00 


IOO 


4 


4320.00 


43.20 











$5000 



4000 



3000 



2000 



1000 











































































































































































































































































































Me 
E 


j.nH 

ngine 


>rizoi 
Cost 


ltal- 












































— E 


san \ 

ngint 


ertic 
Cos 


al 

,s 








































































































































































^ 









































10 



20 



40 



50 
H.P. 



60 



70 



80 



90 



100 



Fig. 15. — Prices of oil engines. 



INTERNAL COMBUSTION MOTORS 



IOI 



Fernold and Orrok in their " Engineering of Power Plants " 
give the cost of producer gas engines as follows : 



TABLE 43 

Producer Gas Engines 

prices, i908-io 





Price f.o.b. Factory 




Size, H.p. 
















Total 


Per h.p. 




20 


$1 ,000 . 00 


$50 . OO 




60 


2,800.00 


46.67 




75 


3,610.00 


48.IO 






f 3,400-00 


42.50 




80 


I 3,250.00 


40.70 






1 3,830.00 


47.90 




85 


| 4,150.00 


48.90 




\ 3,500.00 


41 .80 




100 


4,925.00 


49-25 






[4,950.00 


45.00 




no 


I 4,960.00 


45 10 






( 4,200.00 


37-5o 




130 


5,250.00 


40.40 




135 


6,600 . 00 


48.80 




160 


f 5,500.00 


35 00 






\ 6,100.00 


38.10 




250 


6,650.00 


26.60 






f 12,000.00 


30.00 




400 


\ I2,800'.00 


32.00 




600 


17,400.00 


29.00 




1000 


33,750.00 


33-75 




2000 


64,850.00 


32.43 





79. The prices of 4 cycle Diesel engines run as follows 



TABLE 44 
Diesel Engines, Four Cycle 
1910-12 



Rated 


Price of engine 


Price per h.p. 


60 


$4,620.00 


$77.00 


90 


6,060 . OO 


67 


00 


no 


7,000.00 


63 


50 


150 


9,000.00 


60 


OO 


200 


11,500.00 


57 


50 


250 


14,000.00 


56 


00 


300 


16,500.00 


55 


OO 


400 


21,200.00 


52 


80 


500 


25,800.00 


5i 


60 


750 


37,200.00 


49 


60 


1000 


48,300.00 


48 


30 



102 



BASIC COSTS 



Jp/0,UUUU 










































60,0000 






































/ 




































/' 




g 50,0000 




















































































^40*0000 
o 

| 

* 30,0000 
































































/ 


s 
































































































20,0000 


















































































10,0000 


















































































n 











































200 400 600 800 1000 1200 1400 1600 1800 2000 
H. P. 

Fig. i6. — Prices of producer gas engines. 

The prices of large two-cycle Diesel engines is reported as 
low as $35 per h.p. 

































































































































$50,000 


















































































40,000 

a 


















































































to 

£30,000 

o 


















































































s 
£20,000 


















































































10,000 


















































































n 











































100 200 300 400 500 600 700 800 900 1000 
H.P. 

Fig. 17. — Prices of 4-cycle Diesel engines 



INTERNAL COMBUSTION MOTORS 



103 



80. 










O r-» 





M ■* OV 


M 


8 


"t fOOO 10 


4 !>- VO M 


M 


to 


H 




O H IO M 


CN to H Tf 


to 


10 to C> >o 


Th r-» t^ m 


M 


10 


H 




to cn cn 


rt O CO to 


O 

in 


O CO O to 


tJ- |>. f^ H 




to M 


M 




IO CN IO -sj- 


VO On r^ O 




NtOO v) 


rj- r^ r^ cn 


H 


to w 


w 




10O cn r^. 


0\ CN CN VO 


8 


0\i«5 h 10 


rj- 00 00 cn 




IO H 
#1= 


M 




IO O O <N 


CO t^ O to 




00 


H 4 MO 


IO 00 Ov co 


V© H 






3£ 






O <N O t>- 


Ov «ch 00 to 





M tJ- COO 


to Ov Ov rh 






H 




IO IO N N 


to O 00 vO 


vg 


N t tO N 


vO O O to 


g " 


M M H 




O O to r^ 


O vO 00 Ov 





4 -t 4n 


f^ O H O 




VO M 






#§= 






IO CN CN CN 


v© e* Ov m 


% 


vO tovo 00 


t^ H CN 00 




VO M 






€£= 






to ^" t-- to 


w Ov O Th 


10 


N 1O00 O0 


00 w rt Ov 




VO M 


M M H 




m= 






O to <N f^. 


vO to M VO 





O to M 00 


00 CN tO O 




t^ CN 


M W CN 




m= 






O to O CN 


t^ t-~ CN M 


c? 


* t>ON 


Ov CO t^. CO 




f-» O 






t& 






IO CN CN tN 


Ov H to 





MOHOO 


m m r^ 




r-~ ro 


H M tN CN 




<& 






to O to to 


IO M T^- to 


V> 


NvO N 0> 


CO 00 to CN 




00 to 


M H CN CO 




m 






O IO <N 


M O CN Tf" 


O 
H 


Ov N N H 


JO H Ov 




ft ^ H 


M CN CO CO 




to t^ to 


Tf- O O CM 


»o 


N00 O « 

0> VO M 


OV tO CO to 




M CN Tf IO 






• . . m 


• r • r • '-~ s • 










.5 


: ft : ft*J ft 












'5b 


: * • * d^' 




d 


1 


I 


id 

+J 

U 


J-l 


CD 1T2 CD 3 y 

• S 2 U Td g 

• «J *H bJ O rt 




"8 


£^3 » 


: &i £*&*? 




1 


-SIS 8, 

mil 


S 1 8 S g S -g a 

w <S <u c« J)J2 rt 






W 


W 


PM Ph 


S 


H H U 





Ph 



fc5 



"*3 

.5 h co 
00 - g 
C 22 

2 '5b So 
bO a; bO 
bp co .S 
43 bo 5 



. o 



aft^ 



&8 



„ o 

.9 ^ 



a 
c « ^ 

I -si 
111 
ill* 

„. «*H J" 00 

o o «g o 



T3 Td T3 __, 

Mil 

C/3 t/2 co ^; 

S ^ rf « 

O » CO »i 



Td nd 
£ £ 

CO Cfi 
CO CO 

si «3 

co to CO co c! 

<D <V <D <U o 

^u § y ^o # u .y 
*C "C 'm *C tj 

PL, Ph Ph Ph CD 



104 BASIC COSTS 

In Guldner's " Internal Combustion Motors " (D. Van 
Nostrand Co.) are given very extensive data on gas engines 
from which the data given in Table 45 on illuminating and 
suction gas producer installations are quoted. 

81. The variation of thermal efficiency of gas and oil engines 
with size, of good modern design is as follows: 



TABLE 46 
Thermal Efficiency 

gas and oil engines 

H.p. per cylinder. . . 10 20 30 40 50 60 80 100 725 150 
Full load efficiency, 
percent 20.8 21.8 22.5 22.9 23.2 23.5 24.0 24.3 24.6 24.9 

H.p. per cylinder 200 300 400 500 600 800 1000 

Full load efficiency, per cent. 25.3 26.0 26.4 26.7 27.0 27.5 27.8 

These values correspond to the equation: 

y = *7-3 +3-5 log M 9 (59) 

where y = per cent efficiency, 
and M = the h.p. per cylinder. 

The efficiency of a two cylinder engine is no better than a 
single cylinder engine of half its size, though its regulation is 
usually better. 

82. At fractional loads, the following per cents of full load 
efficiency are obtained : 



TABLE 47 

Per Cent of Full Load Efficiency of Gas and Oil Engines at 
Fractional Loads 

Per cent of full load 100 90 80 70 60 50 40 30 20 10 o 

Per cent of full load efficiency. . . 100 99 96 91 84 75 64 51 36 19 o 

These values correspond to the equation 

y p = F( 2 -F), (60) 

where y p = per cent of full load efficiency, 

and F = per cent of full load expressed as a fraction. 



INTERNAL COMBUSTION MOTORS 
The attendance cost (A) per bo.h.p.h. is 

a _ °-°°5 



105 



VM 



(61) 



where (^4) is expressed as usual in dollars. 

The amount of lubricating oil used is usually 0.005 to 0.007 
pints per br.h.p.h. 













































32 


















































































30 



















































































.2 

£28 

IS 


















































































-4-> 
S 

«26 


















































































2i 


















































































22 


















































































on 











































100 



300 400 



500 
H.X. 



600 



700 



800 900 1000 



Fig. 18. — Full load efficiency of gas and oil engines. 

The maintenance of gas and oil engines is usually expressed 
as between 2 % and 3 % of the first cost of the engine. This is 
not good practice because the first cost of a certain engine in 
use is a constant, while the cost of maintenance varies with 
the price of wages, or, if you please, with the value of the dollar. 
This method could be used provided that the first cost is first 
multiplied by the relative value of the dollar (indexed) before 
the 2% or 3% is taken. Otherwise, it would usually be too 
low, the error greatly increasing with time. 

83. The variation of full load efficiency of Diesel engines 
with size is as follows: 



106 BASIC COSTS 

TABLE 47 

Full Load Efficiency of Diesel Engines 

H.p. per cylinder 10 15 20 25 30 35 40 50 60 70 

Full load efficiency, per 

cent . . 24.0 26.3 27.7 28.5 29.1 29.5 29.8 30.2 30.5 30.7 

H.p. per cylinder 80 100 125 150 200 

Full load efficiency, per 

cent 30.9 31. 1 31.2 31.3 31.4 

These correspond to the equation 

3 ; = -^tan- 1 (^\ (62) 

2.82 V4.2/ 



82 



230 



.28 



£26 



24 













' /** 


. T X 


> r 

1 


, 1 


t -T J- 


1. .... 



20 40 60 80 100 120 140 160 

H.P. 

Fig. 19. — Full load efficiency of Diesel engines. 



180 200 



At fraction loads, the per cent of full load efficiency obtained 

is: 

TABLE 49 

Per Cent of Full Load Efficiency of Diesel Engines at Fractional 

Loads 

Per cent of full load 100 90 80 70 60 50 40 30 20 10 

Per cent of full load 100 99 97 93 88 82 74 65 54 42 

The equation for these values is 

y = 100 — 72(100 — F) 2 (63) 

All the above data is based on the guarantees of manufac- 
turers together with a careful study of the most reliable tests 
reported. The accuracy should be better than can be obtained 
from any single test. 



STANDARD PIPE 



107 



84. We give below prices of various sized pipe as of 191 2, 
inasmuch as we shall have use of such data hereafter. 



TABLE 50 

Standard Pipe 

price (191 2) per ioo feet, random lengths 



S:ze, 
Inches 


Standard 


Extra Strong 


Double Extra Strong 


Flange 


Black 


Galv. 


P.E. 


C &T. 


Plain ends 


Threads 


1 


$1-95 
1 95 
i-95 

2-75 
3.20 

455 
6.25 

7-5o 

9-55 

15-75 

20.70 

28.15 
32.00 
40.00 
44.60 
57- 80 
80.00 

85 .25 
116.00 
1 20 . 00 
i53-oo 
153 00 


$2.75 
2-75 
3-85 
4-45 
6.36 
8.70 

10.40 

i3-5o 

22. 20 
29.10 
38.75 
4405 
54-50 
60.80 
78.80 

i53-oo 
163.00 
221 .00 
227.00 
292.00 
292.00 


$5-25 
5-25 
5-25 
5-25 
5-5o 
8.10 
11 .00 
13.20 

17-75 
29.15 

37-75 

5i-4o 

58.00 

99.40 

110.00 

148.00 

240 . 00 

271 .00 


$5-50 

5-5o 

5 -5o 

5 -5o 

5.80 

8.50 

11.60 

13.90 

18.65 

30.60 

39 65 

54.00 
60.90 








8 

1 








3 








1 

3 


$18.00 
20.00 
25.OO 

35 00 

45.00 

60 . 00 

90.00 

125.00 

165.00 

190.00 










I 

i£ 


$0.06 




if 

2 

2\ 

3 
• 3i 

4 
4* 

5 
6 

7 
8 


0.08 
O.IO 

0-15 

0.20 

0.25 
0.35 
0.55 

o-55 
0.70 
0.85 
1. 00 


$0.90 
1.20 

i-35 
1.50 
1.80 




260 . 00 
360 . 00 


2.20 
2.40 
3-45 
3-90 
4-95 
5-55 






9 
10 






394.00 






1.50 


11 






12 


466 . 00 






2.50 


7-5o 









io8 



BASIC COSTS 



TABLE si 

Standard Casing 

price (191 2) per ioo feet, random lengths 



O.D. 


I.D. 


Wt. per ft. 


Casing 


C-it 


Thread 


Flange 






lb. 










2¥ 


H" 


2.82 


$12.90 


O.IO 


OI5 


i-35 


3 


2| 


3 


45 


15 


57 


O.IO 


O.20 


1.42 


32 


3l 


4 


45 


18 


72 


0.15 


O.25 


1.80 


4 


3! 


5 


56 


22 


68 


0.20 


0.35 


2.16 


4* 


4* 


6 


36 


25 


25 


0.25 


o-55 


2.30 


5 


4l 


7 


80 


31 


59 


0.30 


0.60 


2.50 


6 


51 


10 


46 


37 


00 


0.35 


0.70 


310 


7 


6f 


12 


34 


47 


54 


o-45 


0.85 


4. 60 


8 


7f 


15 


4i 


58 


25 


0.50 


1 .00 


5-25 


10 


9f 


21 


90 


84 


75 


o.75 


1.50 


6.90 


12 


nf 


30-35 


131.00 


1.25 


2.50 


9.00 



$200 

I 

Cm 

©160 

o 
H 

I 120 



80 



40 



























































































































































































£std 


. Iro 


i Pip 


3 












































































































A 


J3td 


, Cas 


ng 
































/ 


/ 


















































-<!v 


i^ood 


Pipe 














X 


x" 









































































6 8 10 12 

Dia. of Pipe, Inches 

Fig. 20. — Prices of pipe. 



11 



16 



18 



20 



WOOD PIPE 



109 



TABLE 52 

Wood Pipe 
(1912) 







PRICE PER 


IOO FEET 


FOR HEADS OF 






Size 
















So' 


100' 


ISO' 


20o' 


250' 


300' 


350' 


400' 


2" 


$9-75 


$10.00 


$10.28 


$11.11 


$11-95 


$12.75 


$14-45 


$16.40 


3 


11 


95 


12.50 


12 


75 


13 


35 


13 


90 


14 


75 


17 


22 


18 


90 


4 


14 


45 


15.OO 


15 


28 


18 


60 


19 


45 


21 


11 


23 


33 


25 


55 


6 


18 


33 


21.40 


23 


33 


25 


28 


27 


5o 


30 


00 


32 


33 


33 


33 


8 


22 


50 


27.78 


31 


40 


35 


55 


38 


60 


4i 


40 


44 


45 


5i 


67 


10 


29 


72 


3583 


40 


83 


46 


95 


50 


85 


55 


83 


59 


17 


68 


33 


12 


33 


61 


43.06 


SO 


55 


56 


11 


63 


06 


70 


00 


81 


40 


90 


00 


14 


44 


00 


60.OO 


























16 


56 


00 


67.OO 


























18 






72.OO 

87.OO 

I02.00 


























20 






























24 


83 


00 



























no 



BASIC COSTS 



TABLE 53 
W. I. Pipe Friction 

FEET PER IOO FEET OF PIPE 



G.P.M. 



5 

IO 

15 

20 
25 
30 
40 

50 

70 

IOO 

120 

150 
17s 

200 

250 
300 
350 

400 

500 
750 

1000 
1250 
1500 

2000 

2500 
3000 
3500 

4000 

4500 
5000 



Size of Pipe in Inches of Diameter 



2 
8 
18 
30 
45 
64 
109 



Harding & Willard: * When slightly rough add 15%. 
When very rough add 30%. 
h * John Wiley & Sons, Inc. 



WOOD AND RIVETED STEEL PIPE 



in 



For large sized pipe the following data are given at Chicago, 
including the cost of laying but less haulage. 

TABLE 54 

Wood Stave Pipe 

cost (191 2) per foot 







Head in Feet 




Size, 








Inches, Diam. 












25 


50 


100 


200 


12 


$0.42 


$0.49 


$0.63 


$0.85 


18 


0.69 


0.80 


1 .02 


1 .46 


24 


0.79 


0.91 


1. 14 


1. 61 


30 


0.96 


1 .12 


1.44 


2.06 


36 


1. 19 


1 .40 


1.82 


2.65 


42 


1 .40 


1.68 


2.23 


3-33 


48 


1 55 


1. 85 


2.46 


3-67 


54 


2.23 


2.62 


3-43 


5.02 


60 


2.85 


3-35 


4-37 


6.40 


66 


3.21 


3-8i 


5.00 


7-38 


72 


365 


4-38 


5-8 3 


8-73 



Wood pipe has made good in practice and has come into 
very general use. The friction of unplaned wood pipe is about 
equal to riveted steel pipe, but that of planed wood pipe is 

given at only 60% to 80% of that of riveted steel pipe. 

/ 

TABLE 55 
Riveted Steel Pipe 
cost (191 2) per foot 



Size, 


Gage Thickness 


inches, diam. 


14 


12 


10 


8 


6 


\ 


l G a 


t 


12 


$0.32 


$0.38 
o.57 


$0.44 
0.65 
0.85 












18 


$0.78 
I.04 

1.27 

i-55 
1. 61 


$0.98 
1.28 
i-59 
i-93 
2.18 
2.48 
2.80 








24 
30 
36 
42 
48 

54 
60 


$i-55 
i-93 
2.30 
2.66 
3-0.3 
3-4i 
3-79 
4-35 
4-52 


$1.99 
2.46 
2.92 

3-37 
3-83 
4.29 

4-75 

5-21 

5-66 








$3 04 
3.58 
4.12 
4.66 

5-21 

5-74 
6.29 
6.83 




































66 












72 























112 



BASIC COSTS 



TABLE 56 

Table of Pipe Friction 

for large riveted steel pipe in feet per ioo feet of pipe 



Cu. Sec. Ft. 


Size of Pipe in Inches of Diameter 


12 


18 


24 


30 


36 


42 


48 


54 


60 


66 


72 


5 
10 


1.6 

4.8 


0.3 

1.0 

1.8 

3-2 




















°-3 
0.5 
0.8 
1.9 


0. 1 
0.2 

0.3 

i-4 
















15 

20 


0.06 
0.2 

0.3 
0.7 














0.05 
0.17 
0.4 
1 .0 












30 

IOO 


0. 1 

0.16 

0.56 














O.08 
O.30 
I .OO 














0.2 
O.66 


O.I 

0.47 
0.84 


0.07 
0.29 
0.59 


200 












300 



































The same authorities give the approximate section of dams 
not used as weirs, and their approximate cost as follows, 
where (h) is the height of the dam in yards. 





TABLE 57 
Approximate Cost of Dams 
1910-12 






Type of dam 


Common 

up stream 

batter, 

hor. to vert. 


Common 
down stream 

batter, 
hor. to vert. 


Approx. sec. 
in sq. yds. 


Approx. cost 
per cu. yd. 


Approx. cost 
per lineal yd. 


Earth 


2 to I 
t.\ to l\ 

2 tO I 

vertical 
0.15 to 1 


3 to I 
i| to I 

2 tO I 

0.8 to I 
0.3 to I 


2.5/z 2 

2. oh 2 

0.4k 2 
o.2$h 2 


$0.50 

2.00 

12.00 
15.00 


$I.2 5 A 2 
2 . 25/f 2 

4 . 00A 2 

4.80/f 2 
3-5°/* 2 


Crib 


Rock fill 

Masonry, straight 

or to 1 

Masonry, arched. 



Tunnels. — On the average, the cost of tunnels including 
timbering and lining is $15 per cubic yard. The costs and 
capacities of tunnels is given below. 



CANALS 



"3 



TABLE 58 

Cost of Tunnels 

1910-12 



Size in ft. 


Sectional 
area, 
sq. ft. 


4X7 


28 


7 X 7- 25 


5° 


IO X 10 


IOO 


12 X 12.5 


I50 


14 X 14- 25 


200 


20 X 25 


500 


30 X 33 


IOOO 



Allowable 

velocity, 

ft. per sec. 



3-6 
10 
10 
10 
10 
10 
10 



Capacity, 
sec. ft. 



IOO 

500 

I,000 

1,500 

2,000 

5, 000 
10,000 



Approx. 
slope per 
1000 ft. 



O.46 
2.0 

1-5 
I . I 
O.g 

0.6 
0.3 



Approx. 
cost per 
lin. ft. 



$16.00 

28.OO 

56.00 

85.OO 

115.00 

280.OO 

500 . OO 



Cost per lin. ft. 
per 100 sec. ft. 



$16 

5 
5 
S 
5 
5 
5 



Canals. — In ordinary earth a velocity of 2 feet per second 
is commonly used. On this basis the following data is given. 



TABLE 59 

Cost of Canals in Ordinary Earth 

1910-12 



Capacity, 


Area Wet, 
Sec. Sq. Ft. 


Depth of 

Water 


Slope in Ft. 
per Mile 


Approx. Cost per Running Ft 


Sec. Ft. 


Low 


High 


SO 


25 


2-5 


4 


$0,375 


$0.75 


IOO 


50 


3-5 


2 


o.75 


i-5° 


200 


IOO 


5 


i-5 


I-50 


3.00 


300 


150 


6 


1 .0 


2.25 


4-5o 


400 


200 


7 


o.75 


3.00 


6.00 


500 


250 


7 


o.75 


3-75 


7 50 


IOOO 


500 


10 


0.50 


7 -5o 


15.00 


1500 


750 


12 


0.50 


11.25 


22.50 


2000 


IOOO 


12 


o-33 


15.00 


30.00 


3000 


1500 


15 


0.25 


22.50 


45 00 



In rock the allowable velocity is 8 feet per second if the canal 
is lined. Under such conditions the following data is given. 
The costs may vary greatly with local conditions. 



ii4 



BASIC COSTS 



TABLE 60 

Cost of Canals in Rock 

1910-12 



Capacity, 


Area Wet, 
Sec. Sq. Ft. 


Water, 
Depth 


Slope in Ft. 
per Mile 


Cost per Running Ft. 


Sec. Ft. 


Low 


High 


SO 


6.25 


2-5 


40 


$0.32 


$1.28 


100 


12.5 


35 


25 


0.63 


2.50 


200 


25.0 


5-o 


16 


i.fiS 


5.00 


300 


37-5 


6.0 


12 


1.87 


7- So 


400 


50.0 


7 


IO 


2.50 


10.00 


500 


62.5 


7 


9 


3-25 


13.00 


1000 


125.0 


10 


6 


6.00 


24.00 


1500 


175.00 


12 


45 


8-75 


35-oo 


2000 


250.0 


2 


3-5 


12.50 


50.00 


3000 


37S-o 


15 


3-o 


18.75 


75.00 



The Reclamation Service gives the following costs of 
excavation. 



TABLE 61 

Cost of Excavations 

1910-12 





Cost per Ctr. Yd. 




Low 


High 


Average 


Plowable with 4 horses 

Plowable with 6 horses 


So. 098 
0. 1225 
0.29 

0-35 
0.60 
0.20 


$l.OO 
2.00 
2.00 
3.00 
5.00 
3.00 


$0.18 
O.30 
O.60 


Indurated material 


Loose rock 


o-75 
2.00 


Solid rock 


Excavation below plane of saturation 

Solid rock below water 


1.80 
4-5o 









The same authorities, the General Electric Company, give 
the following valuable data on hydro-electric installations. 



HYDRO-ELECTRIC INSTALLATIONS 



"5 









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t. per kw. for hydraulic apparatus 
tt. per kw. for generators, excitors 
d switchboards (no transformers) 
ft. per kw. for electric apparatus 

:luding transformers 

1 cu. ft. per kw. for power house 

s transformer 

1 cu. ft. per kw. for power house 
th transformer . 


-7- . m 

§ -.2 


' wT an 




•X* 




. in . 

: § 8 






power house less foundations 

msformers per kw 

power house including foundat 
t no transformers ner kw. . . . 


of hydraulic mach. per kw. . . 

per kw. for generators, excii 

itchboards and cables (no tr 


1 

0. 


m 

S3 S 

a, o 

« & 

8 a* 

22 ° 

C3 O 

•3° 


CU 

1 

•B 
1 


kw 

with foundations and transforme 








■Q m m 

-•> o - 






UO U H H 


U U 


UU 




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U 



n6 BASIC COSTS 

In the above 

Low head, 50 to 200 ft. 
Medium head, 200 to 600 ft. 
High head, 600 ft. and above. 
Small capacity, 200 to 1000 kw. 
Medium capacity, 1000 to 5000 kw. 
Large capacity. 5000 kw. and over. 

The per cent of total costs of various items of construction 
in hydro-electric plants is given as follows: 

Small Low Head Plant 

Hydraulic work not including power house. 55 % 

Power house building 6 % 

Water wheels 9%} Power house fully equipped 45 % 

Electric equipment .30% 

100% 

Medium Low Head Plant 

Dam 43% 1 

Low pressure pipe 20 % \ Hydraulic work less power 

High pressure pipe 3 % J house 66 % 

Power house and machinery 34 % 

Small Medium Head Plant 

Hydraulic work not including power house 

building 76 % 

Power house building 8 % 

Turbines 3-S%\ Power house fully equipped 24% 

Electric equipment 12.5 

Large Medium Head Plant 

Hydraulic work 38 % 

Power house building 10 % } 

Hydraulic machinery 3 1 % i" Power house fully equipped 62% 

Electric equipment 21 % J 

Large High Head Plant 

P am 22 J 1 Hydraulic work not including 

Low pressure pipe 23% power house 61 % 

High pressure pipe 10 % J 

Power house fully equipped 39 % 

85. More complete data on the cost of buildings, excava- 
tions, fills, foundations, bridges, tunnels and the like are given 
in Gillette's "Handbook of Cost Data" (Myron C. Clark 



REFERENCES 117 

Publishing Co.). For similar data on power plants, the 
reader is referred to Fernold and Orrok's " Engineering of 
Power Plants" (McGraw Hill Book Co.), Harding and Wil- 
lard's " Power Plants and Refrigeration " (John Wiley & Sons, 
Inc.), and Guldner's " Internal Combustion Motors" (D. Van 
Nostrand Co.). 

The equations of cost as given by Harding and Willard in 
their "Power Plants and Refrigeration" (John Wiley & Sons, 
Inc., 191 7) are as follows: 



n8 



BASIC COSTS 





a 




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BASIC COSTS 



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124 BASIC COSTS 



PROBLEMS 



1. A 150 h.p. engine has a guaranteed steam consumption of 20$ steam 
per h.p.h. The test shows that the actual consumption is 21.6$. The 
engine operates for 3000 hours per year at full load and costs $4500 installed. 
The steam costs 25 cents per iooo# to produce. The life of the engine is 20 
years. The interest rate is 6%. How much should be deducted from the 
purchase price to compensate the purchaser for the increased cost of operation 
above that guaranteed? 

2. A 300 h.p. engine has a guaranteed steam consumption of 2o# per 
h.p.h. The test shows that the actual consumption is only 18.4$. The 
engine operates for 4000 hours per year at full load, and costs $8500 installed. 
The steam costs 30 cents per iooo# to produce. The life of the engine is 
25 years. Interest rate is 7%. How much should the manufacturer receive 
to compensate for the added economy of service that he has produced? 

3. A power plant consists of two 5000 h.p. engines that consume 2o# 
of steam per h.p.h. Boiler feed temperature 72 , steam pressure i6o# 
and quality 98%. The boilers evaporate 8.5$ of water per pound of coal. 
The engines operate for 3600 hours per year at full load. Interest rate is 
6%. If these engines are removed and sold for $6 per h.p., and new ones 
are installed, having a life of 25 years and using only 12$ of steam per 
h.p.h., how much can be paid for the new engines? Heat costs 18 cents per 
million B.t.u. 

4. A 200 h.p. automatic engine uses 31$ steam per h.p.h. and costs 
$2800. A 200 h.p. Corliss engine uses 25$ steam per h.p.h. and costs $4200. 
Steam costs 25 cents per iooo# to produce. The plant runs for 3000 hours 
per year at full load. Interest rate is 5%. The life of the high speed engine 
is 15 years and for the low speed engine it is 25 years. Which engine is the 
most economical? How much saving will this represent per year? 

6. In problem (4) if the engines are to be used for only 1000 hours per 
year at full load, which engine will be the best? How much saving will 
this represent per year? 

6. In problem (4), for what number of hours of service per year are the 
two engines equal? 

7. A steam plant is to be installed to deliver 1000 kw. at the switchboard. 
A steam engine-generator set can be installed, with a guaranteed economy 
of i5# of steam per i.h.p.h. with a steam pressure of 150$, feed temperature 
180 , quality 97.5%. The mechanical efficiency of the engine is guaranteed 
at 92%, and the generator efficiency at 94%. Or a steam turbo-generator 
can be installed with a guaranteed steam consumption of 20/ per kw.hr. 
with a steam pressure of 150$, feed temperature 150 and 150 of superheat. 
The cost of generating steam in either case is 20 cents per 1,000,000 B.t.u. 



PROBLEMS 125 

The steam engine-generating set costs $23,500. The turbo-generator set 
costs $19,700. Superheaters cost $1800, and boilers cost $8.52 per boiler 
h.p. The attendance cost is 0.1 cent per h.p.h. for the engine-generator 
and 0.06 cent per h.p.h. for the turbo-generator. Interest rate is 5%. 
If the plant is to be run at full load for 6000 hours per year, which outfit 
is the best, and by how much? What is the cost per kw.h. delivered? 

8. If in problem (7), the plant is to be used for 1000 hours per year, 
then which is the best and by how much? What is the cost per kw.h. 
delivered? 

9. In problem (7) for what number of hours of service per year are the 
two units of equal worth? 

10. A boiler plant develops 500 bo.h.p. and uses 3$ coal per bo. h.p. 
The steam pressure is 1 50$ and feed temperature 8o°. A feed water heater 
is added which costs $600 and raises the feed temperature to 180 . The 
plant operates for 4000 hours per year. Coal contains 12,725 B.t.u. per 
pound and costs $4 per ton. Interest rate is 5%. How much saving per 
bo.h.p. h. will be effected? 

11. If in problem (10), instead of a feed- water heater, an economizer is 
added, raising the feed temperature to 300 , allowing 7% for depreciation, 
repair and attendance, how much could we afford to pay for the economizer? 

12. In problem (10), what would be the saving per bo.h.p. if the plant 
were run at full load for only 2000 hours per year? 

13. In problem (11), what could we afford to pay for the economizer, if 
the plant were run at full load for only 1000 hours per year? 

14. In problem (n), how much could we afford to pay for the economizer 
if it were used at full load for only 100 hours per year? For no hours 
per year? 

15. A boiler plant operates at full load for 7000 hours per year. It burns 
50 tons of coal per day costing $4 per ton with operator (A) in charge. 
An analysis of the flue gases shows C0 2 , 5 per cent; O, 15 per cent; and N, 
75 per cent. The coal contains C, 80 per cent, H, 6 per cent, and O, 4 
per cent. Stack temperature 6oo°; boiler room temperature 70 . 

Operator (B) is placed in charge and another analysis taken. This shows 
with the same fuel, C0 2 , 14 per cent; O, 6 per cent; and N, 75 per cent. 
Stack temperature 520 . 

If operator (A) receives $1200 per year, what can we afford to pay operator 
(B)? 

16. If in problem (15), the analysis showed C0 2 , 13 per cent, CO 1 per 
cent, and all other things remained the same, what then could we afford 
to pay operator (B)? 

17. If in problem (15), the analysis showed CO2, 10%, COi, 0.2 per 
cent and O, 10%, with a stack temperature of 550 , while all other 
things remain as before, what then could we afford to pay operator (B)? 



CHAPTER V 
VESTANCES 

Total Vestance. The Time Element. Change Point. Vestances of 
Steam Engines of Various Types at Full and Fractional Loads. 
Vestances at Full and Fractional Loads of Oil Engines. Diesel 
Engines. Induction Motors and Generators. Comparison of Power 
Units of Various Size and Types at Full and Fractional Loads. Ves- 
tances at Full and Fractional Loads of Centrifugal Pumps and of 
Standard and Wood Pipe at Varying Capacities. 

86. As previously shown, the total vestance gives us a basis 
of comparing two or more units, or two or more systems that 
are designed to render similar service but each differing from 
the other in one or more of their cost elements, such as first 
cost, life, efficiency, attendance and the like. The question 
that both the manufacturer (producer) and user must face con- 
tinually is one of economy. The user desires to purchase that 
unit which will give him the desired service at the least ulti- 
mate cost. The manufacturer's problem leads to the same 
end, namely to produce a unit which will represent the best 
possible investment for the purchaser. But the purchaser's 
needs are variable. What may be best for one, may be a very 
poor investment for another purchaser, whose conditions may 
be entirely different. It is therefore necessary to produce 
various classes and types of apparatus, to meet these variable 
conditions so that a best investment for each condition may be 
offered. That is the problem in general. 

Specifically the producer may face such a problem as this: 
a more efficient engine can be built, but at a greater first cost. 
If produced, will this engine represent a better investment 
than the unit now being produced and will it sell? Whether 
the new engine will represent a better investment for any 
given service, can be readily determined by comparing the total 
vestances. If that of the new engine is the smaller, greater 

126 



THE TIME ELEMENT 127 

financial efficiency has been attained. Whether the engine 
will sell, depends on whether it is really a better investment or 
not, and by how much, and second whether we can prove to 
the user that it is a better investment. Of course if we can 
prove it for ourselves, we can prove it to the user, but when 
it comes to a matter of guess, one man's guess is quite as good 
as another's. At all times we must bear in mind that we do 
not purchase the machine, but the service that it will render 
us. Manufacturers therefore produce service in just the same 
sense that the so-called public utilities do. 

87. The Time Element in Vestance.— The total vestance 
depends upon the amount of use that is made of the apparatus. 
This may be shown as follows: 

The total vestance is 

r -r R , 

neglecting insurance and taxes, where 
C = the first cost, 
T = the term factor, 
R = the interest rate, 
and A = the annual cost of operation. 

The annual cost of operation (^4) varies with the number of 
hours of use per year and the cost of operation per hour. If 
a = the mean operating cost per hour of the entire 

unit or system and 
N = the number of hours of use per year, then 
A = aN 
and the total vestance becomes 

T/ C a N fK >. 

(^ T R 

Dividing equation (i£) through by (A 7 ) gives the vestance 
per hour V h of service or 

v ^Yn + A r <*> 

By equation (65) it is evident that as (N), the number of 
hours of use increases, the vestance per hour decreases. The 



128 VESTANCES 

maximum value that (A) can attain is 8760, the total num- 
ber of hours in a year. For this value of N f the vestance per 
hour of service is a minimum. When, on the other hand, ( A) 
decreases, the vestance per hour increases, becoming infinite 
when N = o. 

Evidently then as ( N) becomes smaller, and the term (. ) 

becomes larger, the constant term ( — ) becomes of less and 

less consequence. We can interpret this for practical use by 
saying that where the unit is but little used per year, the first 
cost is of primary consideration, and the operating cost per 
hour (efficiency and energy cost) of secondary importance 
only, while for many hours of operation per year conditions 
are reversed. That is, when the apparatus is much used, we 
can afford to buy expensive equipment in order to get low 
cost of operation, but where little used, cheap apparatus of 
comparatively low efficiency is the best. 

88. Valuance. — If we take the reciprocal of the vestance 
per hour (V h ), we get by equation (65) 

TT J TRN (fA 

U = V h = CR+TaN (66) 

We shall call (U) the valuance. It is a measure of the 
worth of the unit. When (A), the number of hours of use, is 
zero, the valuance is zero, i.e., the unit is worthless (useless). 
The valuance increases as ( N) increases, becoming a maxi- 
mum when A 7 = 8760, i.e., the apparatus is worth most (most 
useful) when it is used continually. 

89. We shall now determine the vestances of various types 
of equipment for a normal year of 3000 hours, assuming a 
constant load during the entire period. In this way we can 
get a first approximation of the comparative values of such 
equipment. Further on, we shall consider more variable and 
exact conditions. 

90. The Steam Engine. — From the previous chapter, we 
find that the first cost of a 100 h.p. low speed compound steam 



THE STEAM ENGINE 129 

engine is $3300. The efficiency with dry saturated steam at 
ioo# initial steam pressure is 12 %. The cost of oil, attendance, 
etc., is about $0.05 per hour for this size of engine, and the 
maintenance about $0.02 per hour. The life of low speed 
compound engines is 25 years. 

If the interest rate is 5%, then by Table 2, the term factor 
is 0.70469, so that the depreciation vestance is 

V d = 33 °° = $4682.90. 
0.70469 

If the coal used for fuel contains 13,743 B.t.u. per pound, 
its energy content in h.p.h. is 

iM13 = h h< 
2545 F 

If this coal costs $2 per ton, or $0,001 per pound, then its 
cost per h.p.h. is 

0.001 -J- 5.4 = $0.000185. 

If the combined efficiency of the boiler, furnace, and grate 
is 70 %, then the cost per h.p.h. of heat energy delivered to the 

engine is 

0.000185 -f- 0.7 = $0.000264. 

To this we must add the other costs of producing steam, 
amounting to, let us say, $0.000236, giving a total cost of the 
heat delivered to the engine of 

0.000264 + 0.000236 = $0,005 P er h.p.h. 
Since the efficiency of the engine is 12 %, the cost of the heat 
per i.h.p. is 

0.0005 -^ 0.12 = $0.00417, 

and for 100 h.p. this amounts to 

0.00417 X 100 = $0,417. 
To this must be added the other operating costs of $0.07, 
giving a total of 

0.417 + 0.07 = $0,487 per hr. 

For a year of 3000 hours, this amounts to 
0.487 X 3000 = $1461, 
giving an operating vestance at 5 % interest, of 
1461 -T- 0.05 = $29,220. 



i 3 o VESTANCES 

Since the depreciation vestance is $4682, the total vestance 
is 

V t = 4682 + 29,220 = $33,902, 

or $339.02 per i.h.p. 

Note that the operating vestance is over six times as great 
as the depreciation vestance. In other words the first cost is 
of relatively small importance as compared with the costs of 
operation, a matter not always fully appreciated. 

91. As compared with the above, let us determine the total 
vestance of the above engine when the initial pressure of the 
steam is 2oo#, other things remaining the same. 
Since for compound engines, the efficiency is 
y = 0.0675 (E - 145), 
and the steam temperature (£) at 2oo# is 3 88°, we find that 

y = 0.0675 (388 - 145) = 16.4% 
instead of 12% previously taken. 
The heat will cost therefore 

0.0005 -4- 0.164 = $0.00305 per h.p.h., 
or 0.00305 X 100 = $0,305 per hr., 

giving a total cost per hour of 

0.305 + 0.07 = $0,375, 
or 0.375 X 3000 = $1125 per normal year. 

The operating vestance is then 

1125 + 0.05 = $22,500, 
giving a total vestance of 

22,500 + 4682 = $27,182. 
This shows a reduction of 

33,902 — 27,182 = $6720, 
or $67.20 per h.p. 

If we add to the 2oo# initial gage pressure 200 of super- 
heat, then the efficiency will be increased by 20 %, amounting 
to 

1.20 X 16.4 = 19.68%. 

The cost of heat per hour will then be 

0.0005 X 100 q 

2—^ = $0,254, 

0.1c* 



THE STEAM ENGINE 131 

and the total cost per hour will be 

0.254 + 0.07 = $0,324, 
or 0.324 X 3000 = $972, 

giving an operating vestance of 

97 2 T 0-05 = $i944o, 

and a total vestance of 

4682 + 19,440 = $24,122. 
This represents a saving of 

27,182 — 24,122 = $3060, 
as compared with the same unit, not using superheat but the 
same pressure, and 

33,902 — 24,122 = $9780, 
as compared with the engine using dry saturated steam at 
ioo# initial pressure. 

The per cent saving by use of 2oo# pressure as compared 
with ioo# initial gage pressure is 

6720 -4- 33,902 = 20%. 
The saving in total vestance by the use of 200/ pressure 
together with 200 superheat as compared with the use of ioo# 
dry steam pressure is 

978o -f- 33.902 = 28.8%. 
Note the enormous saving especially with the use of higher 
pressures. Modern practice tends towards continually higher 
pressures. That this is good practice is too clearly shown 
above to need further emphasis. 

92. Let us compare with the above the total vestance of a 
1000 h.p. compound engine, the life, rate of interest and cost 
of coal being assumed the same as that for the 100 h.p. 
engine. 

For this engine the first cost is $18.60 per h.p. and the de- 
preciation vestance is then 

18.60 -T- 0.70469 = $26,394 per i.h.p. 
At ioo# initial gauge pressure, and dry saturated steam, the 
efficiency of the engine is 14.7%. The heat cost per h.p.h. 
will then be 

0.0005 -f- 0.147 = $0.0034. 



132 VESTANCES 

Allowing $0.0006 as the cost of attendance, maintenance, 
etc., per h.p.h., the total operating cost per h.p.h. will be 

0.0034 + 0.0006 = $0,004, 
or 0.004 X 3000 = $12 per h.p. year. 

The operating vestance is then 

12 -T- 0.05 = $240 per h.p., 
and the total vestance per h.p. year is 

240 + 26.394 = $266,394, 
as compared with a total vestance of 339.02 per h.p.h. for the 
100 h.p. engine. This gives a difference of 
339.02 - 266.394 = $72,626. 
The per cent saving as compared with the 100 h.p. engine of 
72.626 -=- 339.02 = 21.4%. 

This is of course not a clear saving due to the greater dis- 
tribution area, and corresponding greater distribution cost for 
the larger plant. In fact the vestance of this increased cost 
of distribution may, if the system is too large, more than off- 
set the decreased vestance of the engine itself. 

93. We shall now determine the total vestance of a 100 h.p. 
low speed simple noncondensing engine operating on dry 
saturated steam at ioo# initial gage pressure, the life, cost 
of coal, and interest rate remaining the same as above. 

The first cost of such an engine is $21.50 per h.p. The 
depreciation vestance is therefore 

21.50 -4- 0.70469 = $30.51 per h.p. 

This engine will use 42% more steam than a compound 
condensing engine. The efficiency of the simple noncon- 
densing engine of 100 i.h.p. will therefore be 
12 -=- 1.42 = 8.45%. 

The cost of the heat energy per h.p.h. will therefore be 
0.0005 -f- 0.0845 = $0.0059. 

Allowing as before $0.0007 for the other operating costs, we 
get a total of 

0.0059 + 0.0007 = $0.0066 per h.p.h., 
or 0.0066 X 3000 = $19.80 per h.p. year. 



THE STEAM ENGINE 133 

The operating vestance is then 

19.80 -7- 0.05 = $396, 
and the total vestance is 

396 + 30.51 = $426.51, 
as compared with 339.02 for the 100 h.p. condensing engine. 
The difference is 

426.51 - 339.02 = $87.49, 
or 87.49 -f- 339.02 = 26.8%. 

94. For a 100 i.h.p. simple noncondensing high speed en- 
gine operating on dry saturated steam at ioo# initial gage 
pressure, the first cost is $12.75 per h.p. but the life now is 
only 15 years instead of 25 years for the low speed engines. 
The depreciation vestance is then 

12.75 ■* 0.51897 = $24.58. 
Since the simple high speed noncondensing engine will use 
64% more steam than a compound condensing engine and the 
efficiency of a 100 i.h.p. engine of the latter type is 12%, the 
efficiency of the high speed engine will be 

12 ^ 1.64 = 7.3%. 
The cost of the heat energy per h.p.h. is then 
0.0005 -T- 0.073 = $0.00685. 
Adding thereto the other operating cost per h.p.h. we get for 
the total operating cost per h.p.h. 

0.00685 + 0.0007 = $o- 00 755> 
or 0.00755 X 3000 = $22.65 P er h-P- year. 

The operating vestance is therefore 

22.65 * 0.05 = $453, 
and the total vestance is 

453.00 + 24.58 = $477-58, 
as compared again with 339.02 for the compound engine. 
This gives a difference of 

477-58 - 339-02 = $138.56, 
or 138.56 -r 339.02 = 41%. 

95. In order to bring out the differences more clearly, we tabu- 
late the vestances of the three classes of 100 i.h.p. engines below. 



134 



VESTANCES 



TABLE 64 

Vestances 
ioo i.h.p. engines with dry saturated steam at ioo# 


G.P. 


Engine 


Vd 


Va 


V 


Per cent Vt. 


Va 
Vd 


A 

B 
C 


46.82 
3051 
24-58 


292.20 
396 . OO 

453- 00 


339.02 
426.51 

477-58 


1 .OOO 
I.268 
1. 410 


6.23 
I3.OO 
18.4O 



Here engine A = low speed compound condensing; B = low 
speed simple noncondensing; and C = high speed simple non- 
condensing. 

Note now that between engines A and B we save $16.31 
in depreciation vestance at a cost of $103.80 in operating ves- 
tances, and between A and C we save $22.24 in deprecia- 
tion vestance. Neither of these represent very good bargains. 
Yet how often are such bargains entered into ! 

96. We tabulate below the vestances of steam engines, 
based on 5% interest, $2.50 cost per ton of coal of fuel value 
as above, i.e., 5.4 h.p.h. per pound and 25 years' life for low 
speed engines, and 15 years for high speed engines, all as taken 
above. The effect of increased cost of operation is to increase 
the operating vestance by a similar amount. Thus if coal were 
$5 per ton and other operating costs increased in like pro- 
portion, the operating vestance would be doubled, amounting 
to some $400 increase. If on the other hand the first cost is 
doubled, the depreciation vestance will be doubled, but this 
will amount only to some $30. Evidently then the effect 
of increased prices is to make efficiency relatively more im- 
portant and first cost, though itself increased, of relatively less 
importance. 

The effect of increased interest rate is to decrease the 
vestances. 



THE STEAM ENGINE 



135 



TABLE 65 

Vestances per H.p. of Compound Condensing Low Speed 
Steam Engines at Full Load 







Vestances 




Ratio 


Size, 










I.H.P. 








Va 

Vd 




Depreciation Vd 


Operating Va 


Total V 


IOO 


$46,829 


$292.00 


$338.83 


7.2 


200 


35-476 


271 .00 


306 . 48 


8.7 


3OO 


31.688 


262.00 


293 . 69 


9-3 


4OO 


29 . 800 


256.00 


285.80 


9.6 


500 


28.736 


251 . 20 


279.94 


9-7 


60O 


27.913 


247.40 


275-51 


9.8 


700 


27.572 


244.10 


271 .67 


9.9 


800 


26.962 


242.OO 


269 . OO 


10. 


9OO 


26.600 


241 .10 


267.70 


10. 


IOOO 


26.394 


240 . OO 


266 . 39 


10. 1 


I500 


25.642 


236.00 


261 .64 


10.2 


2000 


25.259 


234.00 


259.26 


10.2 













































$400 


















































































o;aoo 

X 




























































1 


otal 


Vesta 


nee- 


* 














u 
"200 

I 

>100 














Open 


ting 


Vest 


mce- 


3 — 


































































































































































b 


spree 


iatioi 


1 "Vestaacd 


■^v 










A 






















| 













200 400 



600 800 1000 1200 1400 1600 1800 2000 
H.P. 



Fig. 21. — Vestances per h.p. of compound low speed condensing 
steam engines. 



136 



VESTANCES 



TABLE 66 

Vestances per H.p. of Simpie Low Speed Noncondensing 
Steam Engines 





Vestance at Full Load 




Ratio 


Size, 








I.H.P* 








Va 

v d 




Depreciation Vd 


Operating Va 


Total V 


100 


$3° -5IO 


$398 • OO 


$428.51 


13. 1 


125 


28.381 


386 . OO 


414 


38 


13 


7 


150 


26.962 


382.OO 


408 


96 


14 


2 


175 


25.898 


374.OO 


399 


90 


14 


4 


200 


25.188 


368 . OO 


393 


19 


14 


6 


250 


24.124 


362.OO 


386 


12 


15 





300 


23-4I5 


358-00 


381 


4i 


15 


3 


3SO 


22.918 


354 -oo 


37^ 


92 


15 


4 


400 


22.521 


350.00 


372 


52 


15 


5 


450 


22.237 


346 . 00 


368 


24 


15 


6 


500 


21 .996 


342.00 


364 


00 


15 


5 



TABLE 67 ' 

Vestances per H.p. of Simple High-speed Noncondensing 
Steam Engines 





Vestance at Full Load 




Ratio 


Size, 










I.h.p. 








Va 

Vd 




Depreciation Vd 


Operating Va 


Total V 


20 


I49-3I3 


$874. OO 


$923-3I 


17.7 


30 


36 


328 


618.OO 


654 


33 


17.0 


40 


29 


800 


550.00 


579 


80 


18.5 


50 


25 


898 


5°3 OO 


528 


90 


19.4 


75 


20 


718 


474.OO 


494 


72 


22.9 


100 


18 


093 


454.OO 


472 


09 


25.0 


150 


15 


462 


434.OO 


449 


46 


28.5 


200 


14 


191 


418.OO 


432 


19 


29.6 


250 


13 


339 


412.00 


425 


34 


31-4 



THE STEAM ENGINE 



137 













































$1000 










































\ 








































c 












































\ 






































1 
£600 

1 

400 




\ 








































\ 


^ , 
















































yS 


mple 


High- speed Non-condensing 


Engi 


aes 
sonde 


nsingJEngines- 








_Simple_Low-speed._KoiM 




















1 ' 

2ompoun< 


Low 


-spee 


d Co 


adem 


ing I 


]ngines 






200 



























































































































50 100 150 200 250 300 350 400 450 500 

H. P. 

Fig. 22. — Total vestances per h»p. of steam engines. 

A comparison of these tables shows that the total vestance 
decreases with increased size of unit. The decrease is rapid 
for units less than ioo h.p., rather small between ioo and 
500 h.p., and almost insignificant beyond that. Again the 
approximate mean vestance ratio is 10 for compound con- 
densing engines, 15 for simple low speed noncondensing en- 
gines and about 25 for simple high speed noncondensing 
engines. That is as the first cost is reduced, the relative 
amount of money that must be spent for operation is enor- 
mously increased. In as much as the vestance is proportional 
to the total amount of labor that it takes to produce a given 
service, including not only the labor of operation but of the 
construction of the units and the system, it appears that in 
cheap layouts, too little effort has been spent in the design 
and construction of the units and system and too much must 
therefore be spent in operation. Less total labor would be 
required if it were used in the right place, in properly prepar- 
ing for the rendition of the service first, before rendering it. 
On the other hand, too much may be spent on preparation. 

97. The above tables were figured on the basis of a normal 
year of 3000 hours of use. If the number of hours of use are 
changed, then the cost of operation is changed, since the latter 



138 VESTANCES 

is in direct proportion to the hours of use. But the deprecia- 
tion vestance remains constant. We have then, that 

V = V D + V a 'N, (67) 

where 

Va = the operating vestance per hour, 
and N = the hours of use, 

Vd = the depreciation vestance, 
and V = the total vestance. 

From the above, it is evident that the vestance per hour is 

V h =?f+V a > (68) 

Example 27. — Determine the total vestance per hour of a 
500 h.p. compound condensing steam engine for a variable 
number of hours of use per year. 

Solution: By Table 65, we have 
V = $28,736, 
and V a = $251.20 for N = 3000 hrs. 

Then V a ' = ^^ = $0.08373. 

3000 

Therefore the total vestance per hour is 

+ 0.08373. 



V h 

$0.08948 

0.1 1 247 

0.3711 

28.82 

00 

The total vestance of this engine is 

V = 28.736 + 0.08373 N. 

Example 28. — Determine likewise the total vestance per 
h.p. for (a) a 100 h.p. compound condensing engine, (b) for a 
100 h.p. simple low speed noncondensing engine and (c) for 
a 100 h.p. simple high speed noncondensing engine. 





Vh' 


— 


N * 


this, 


we tabulate 

N 
5000 hrs. 
1000 hrs. 
100 hrs. 
1 hr. 
hr. 


as 


follows: 



THE STEAM ENGINE 



139 



Solution: (a) In this case, we have by table 64 



VI' 



or 

so that 



46.829 292.00 

N 3000 

Jr ,, 46.829 

Vh = ^-^ + 0-09733, 



F" = 46.829 + 0.09733 # (0) 

(6) In this case we have by Table 65 

V h _-_+_ 



3000 
or TV" = 25^9 + 0.13267, 

so that V" = 30.510 + 0.13267 N (b) 

(c) In this case, we have by Table 66 

7 fc "" =^3 + aisi33> 

or 7"" = 18.093 + 0.15133 N (c) 

Plotting these equations (a), (6), and (c) gives straight lines 

as shown in Fig. 23. 

It may be that for a certain number of hours of use per 



$S8(J 






































/ 


s 


240 


































, • 


/ 


































j* 


u 










200 




























*j5 


lX 




*#£ 






























c 


# 




^ 


U5 











J 160 

en 


























*; 


rt 
















































g^S 


tf* 6 ,. 




> 

3 120 




























Dot^ 


OM" -^ 


Co»| 


Ve£^ 












































80 














































B 


C 


































40 




^ 


^ 




































-^ 



















































































200 



400 



600 



800 1000 1200 1400 
Hours of Use per Year 



1600 1800 2000 



Fig. 23. — Variation of total vestance with hours of use for 100 h.p. 
steam engines. 



i4o VESTANCES 

year, that the total vestance for the compound and simple 
low speed engines are equal. In such a case then 

V" = V", 
or 46.829 + 0.09733 ^1 = 3°-5 10 + 0.13267 Ni, 

whence Ni = 461. 1 hrs. 

So again the total vestance per year ( Vsl) for* the simple 
slow speed engine is 

V S l = 3°-5 10 + 0.13267^, 
and for the simple high speed engine it is 

V S h = 18.093 + 0.15133^. 
These two become equal when 

18.093 + 0.15133 iV 2 = 3°-5 10 + 0.13267N2, 
that is when N2 = 661.5 nrs - 

So again the vestance of the compound engines equals that 
of the simple high speed engine when 

18.093 + 0.15133N3 = 46.829 + 0.09733^3, 
or ZV 3 = 532.1 hrs. 

98. Change Points. — The meaning of these three points 
is clearly shown in Fig. 23 where the total vestance per horse 
power year has been plotted against the hours of use. The 
equations for vestance in terms of hours of use give straight 
lines as shown. 

From this it is evident that when the hours of use are less 
than 46 1. 1, then the simple low speed engine, in spite of its 
lower efficiency, is more economical in use, i.e., has better 
financial efficiency than the compound engine. When used for 
more than 461.1 hours, however, the compound engine is best. 

So also when the hours of use are less than 532.1, the simple 
high speed engine has better financial efficiency than the com- 
pound engine. But this number of hours is greater than that 
of the simple low speed engine. So then for more than 532.1 
hours per year use the compound engine; for less than this 
number of hours use the simple high speed engine and do not 
use the simple low speed engine at all. This is for the reason 
that for no hours of use during the year is it best, being always 
exceeded, either by the compound or simple high speed engine. 



CHANGE POINTS 141 

We shall call the point of equality as 2Vi, A 2 , and A3 above 
the Change Points. 

99. For the 200 i.h.p. engines, we have a total vestance for 
the compound engine of 

Vc = 35476 + 0.09033^ (a) 

for the simple low speed engine of 

V SL = 25.188+ 0.12267^, (b) 

and for the simple high speed engine of 

V S h = 14.191 + °.i3733N (c) 

Equating equations (1) and (2) gives 

25.188 + 0.12267^1 = 35.476 + 0.09033N1, 
so that Ni = 318 hrs., 

and so also equating equations (1) and (3) gives 

14.191 + 0.13733 N 2 = 35-47 6 + 0.09033N2 
so that N 2 = 452 hrs. 

Again the value of the change point is greater for the high 
speed than for the low speed engine with the compound engine. 
So again we find that the simple low speed engine possesses 
no period of superiority and should therefore under the condi- 
tions assumed not be used for this size. For this size and these 
conditions, the compound engine should be used, if the period of 
use is over 452 hours, and when the period of use is less than 
this number of hours, then the simple high speed engine gives 
the highest financial efficiency. 

100. So far we have discussed only the condition of an 
engine operating at full load. If instead the engine is used 
at a constant fractional load, the efficiency will be decreased 
and thereby the cost of heat will be increased. The other 
operating costs such as attendance and the like will be 
changed to correspond to the actual load carried. The depre- 
ciation vestance will change in inverse proportion to the per 
cent of full load carried. Thus at half load it will be doubled; 
at one and one-quarter load it will be four-fifths as great, etc. 
This is evident since, for example, at half load we have pur- 
chased two horse power to get one. 

We give below the vestances for each of the three types of 
engines at 1.25, 0.75, 0.50 and 0.25 of full load, the per cents 



142 



VESTANCES 



of full load efficiency at fractional loads being assumed the 
same for simple low speed engines for want of authentic data 
on the former. The vestances are per actual horse power 
developed and not per rated horse power. A normal year of 
3000 hours is again assumed. 

TABLE 68 

Vestances per H.p of Low Speed Compound Condensing 

Steam Engines at Fractional Loads 

dry saturated steam at ioo# g.p. 



Rated 


Total Vestance at Loads or 


H.p. 














125% 


100% 


75% 


50% 


25% 


IOO 


$345-5 


$338.8 


$361.4 


$411.6 


$544-3 


200 


3*3 


4 


306 


5 


321.3 


361.0 


469 


9 


300 


301 


4 


293 


7 


3°8-3 


345-4 


444 


8 


400 


291 


8 


285 


8 


298.3 


334-2 


424 





500 


284 


9 


280 





291.8 


326.4 


415 


6 


600 


280 


3 


275 


5 


286.6 


321.8 


407 


6 


700 


278 


2 


271 


7 


283.0 


318.5 


403 


7 


800 


276 


6 


269 





281. 1 


315-4 


400 





900 


275 





267 


7 


279.0 


3i3-i 


397 


6 


IOOO 


274 





266 


6 


277.6 


3H-4 


395 


6 


1500 


270 


5 


261 


6 


273.2 


306.0 


387 


2 


2000 


267 


5 


259 


3 


270.5 


302.2 


383 


8 



200 



Fig. 24. 



400 



600 



800 



1000 
H. P. 



1200 













































$600 


















































































500 

1 

$ 400 


















































































i 


V<- 
























254 














00 




> 


X v 




































3 ■ 

gsoo 


























t 


^-50 


#Lo 


A 






















-*= 










•75$|Loa 


i— 










200 




















126jt 


Load 




K 


-100 


*L< 


ad 








































































































































































_ 



1400 1600 1800 2000 



Vestances per h.p. of low speed compound condensing 
steam engines. 



CHANGE POINTS 



143 



TABLE 69 

Vestances per H.p. of Simple Low Speed Noncondensing 
Steam Engines at Fractional Loads 

dry saturated steam at ioo# 



Rated 


Total Vestances at Loads of 


H.p. 


125% 


100% 


75% 


50% 


25% 


100 


$442.2 


$429 -5 


$442.6 


$496 . 4 


$609.0 


125 


433 


1 


416 


2 


431 


9 


478 


2 


590.6 


I50 


424 


4 


406 


4 


422 





466 


4 


574 .0 


175 


417 


1 


398 


9 


412 


9 


456 


8 


562.6 


200 


410 


2 


393 


6 


40S 


9 


448 


4 


554-8 


250 


401 


9 


386 


1 


395 


8 


440 





543 


300 


396 


3 


379 


4 


39° 


1 


434 


2 


531-6 


350 


39o 


7 


374 


7 


384 


2 


427 


8 


524.6 


400 


384 


8 


369 


9 


379 


6 


421 


6 


5i7-o 


450 


380 


6 


364 


6 


375 


7 


416 


4 


510.4 


500 


375 


6 


360 


6 


37o 


9 


412 





505.4 



TABLE 70 

Vestances per H.p. of Simple High Speed Noncondenslng 
Steam Engines at Fractional Loads 

dry saturated steam at ioo g.p. 



Rated 




Vestances at Loads 


OF 




H.p. 


125% 


100% 


75% 


50% 


25% 


20 


$742-3 


$701.3 


$725-6 


$806.6 


$1005 . 2 


30 


647 





618 


3 


638 


5 


706.6 


867.2 


40 


587 


8 


56i 


8 


579 


8 


639.6 


780.2 


5o 


555 


7 


529 


9 


544 


5 


602.8 


731.6 


75 


5i8 


5 


494 


7 


506 


8 


558.0 


673.6 


IOO 


499 


1 


474 


1 


485 


2 


532.4 


6 37-4 


150 


474 


8 


449 


8 


460 


2 


504.O 


602.4 


200 


457 


4 


432 


8 


442 


3 


487.4 


574-8 


250 


448.4 


425-9 


434- 1 


474-4 


560.2 



144 



VESTANCES 













































$1000 












































\ 






































900 




1 


\ 








































\ 




































B 800 

u 

S 






\ 






































\ 






































J* 

CO 




ty 


\ 






































s 


%> 




































> 






\ 


\ 


















25£ 


Load 














O 






\ 


\ 
































500 


























50 # 


Loac 
























75 


% Lo 


ad^ 














>ad~t 
























4001 




















100 # L< 


ad^ 

















25 



50 



75 



100 



125 
H. P. 



150 



175 



200 



225 



250 



Fig. 25. — Vestances per h.p. of simple high speed noncondensing 
engines at fractional loads. 

101. It will be noticed that especially in the simple low 
speed noncondensing engines the total vestance per h.p. at 
three-quarters load is but little greater than at full load. The 
total vestance at half and one and one-quarter loads is not 
very great as compared with that at full load, while that at 
one-fourth load is comparatively very great. The latter is 
evidently then a very uneconomical load at which to operate. 

Since the above tables are based on 3000 hours of opera- 
tion, we can again get the variation of total vestance with the 
number of hours of operation per year, and thus determine 
the change point between two different types of engines at frac- 
tional loads. 

Example 29. — Determine the change point between a 
100 h.p. compound engine running at full load and a 200 h.p. 
compound engine running at half load. 

Solution: The depreciation vestance of a 100 h.p. engine at 
full load is $46.80, and the operating vestance is $292 per year 
or 292 -T- 3000 = $0.09733 per hr. 

The total vestance ( V) is then 

V = 46.80 + 0.09733 N. 



CHANGE POINTS 145 

For the 200 h.p. compound engine operating at half load, 
the depreciation vestance is 

35.5 X 2 = $71.00 per h.p., 

and the operating vestance is the difference between the total 
and depreciation vestances or 

361 - 71 = $290 per yr., 
or 290 -f- 3000 = $0.09667 per hr. 

The total vestance is then 

F200 = 7 1 + 0.09667 N. 

The change point is obtained by equating 

F100 and F 20 o 
or 46.80 + 0.09733^1 = 71 + 0.09667 2Vi, 

or Ni = 36,120 hrs. 

Since there are only 8760 hours in a year, the change point 
does not come within the year. At all times then the 100 
h.p. engine is the most economical to use at 100% load, the 
decreased operating vestance of the 200 h.p. engine being 
insufficient to offset the increased depreciation vestance over 
that of the smaller engine. 

Example 30. — Determine the change point of a 250 h.p. 
simple high speed engine running at full load and a 200 h.p. 
engine of the same type running at 25% overload. 

Solution: In this case we have for the 250 h.p. engine 
V D ' = $13.30, 
and VJ = 412.60 per yr., 

or 412.6 4- 3000 = $0.13753 P er nr -j 

so that VI = 13.30 + 0.13753 N. 

So also for the 200 h.p. engine at 125% load 

V D " = 14-2 4- 1.25 = $11.40, 
and V" = 457-4 — 1 1. .4 = $446 per yr., 

or 446 4- 3000 = $0.14867 per hr., 

so that V" = 11.4 + 0.14867 N. 



i 4 6 VESTANCES 

Equating V{ and V" gives 

11.4 + 0.14867 #1 = 13.3 + 0.13753 Ni 
so that Ni = 170 hrs. 

This shows that if an overload of 25% is to be carried less 
than 170 hours per year, it can be done more economically 
with the use of the 200 h.p. engine at 25% overload than by 
using a 250 h.p. engine at full load. 

102. Oil Engines. — We give in Table 71 the vestances per 
h.p. of oil engines per normal year of 3000 hours, at 5% 
interest rate. These are also based on a fuel cost of 0.51 1 
cent per pound of oil having a heat value of 18,600 B.t.u. 
per pound. This is equivalent to a heat value of 7.3 h.p.h. 
per pound and a cost of heat energy of 0.07 cent per h.p.h. 
The life of the engines is taken as 15 years, although this 
is usually exceeded. 



OIL ENGINES 



147 



TABLE 71 

Vestances per H.p. of Oil Engines 



Type 



i cy. horiz. 



1 cy. vert. 



1 cy. horiz. 



1 cy. horiz. 



1 cy. vert. 



1 cy. horiz. 



1 cy. vert. 



1 cy. horiz. 



1 cy. horiz. 



1 cy. vert. 



1 cy. horiz. 



Loads 



100% 



V D $174.00 

V a 44O. OC 



V t $614.00 

Vd $174.00 

V a 44O.OO 



Vt $614.00 

V D .$174.00 

Va 394 50 

V t $568.50 

V D $174.00 

V a 367.00 



75% 



V t $541.00 

Vd $130.50 

V a 367.00 



Vt $497-5o 

V D $161.00 

V a 332.00 



Vt •• . .$493.00 

Vd $116.00 

V a 332.00 



Vt $448.00 

Vd $151.00 

V a 312.00 

Vt $463.00 

Vd $130.50 

V a 287.OO 

Vt $417-5° 

Vd $116.00 

V a 287.OO 

Vt $403-00 

Vd $116.00 

V a 265.50 



V t $381.50 



$231.50 

485 • 00 
$716.50 

$231.50 

485-00 

$716.50 

$231.50 

437 00 

$668.50 

$231.50 

405 • 00 

$636.50 

$184.50 

405 • 00 

$589-50 

$215.00 
367 - 00 

$582.00 

$154-50 

367-00 

$521.50 

$202.50 

342.QO 

$544-5o 



$184.50 

3i4-oo 

$498.50 

$i54-5o 

3*4-00 

$468 . 50 

$i54-5o 
289.50 



50% 



$444 . 00 



$348.00 

607 . 00 

$955.00 

$348 . 00 

607 . 00 

$955 00 

$248 . 00 

540 00 

$788.00 

$248 . 00 

500 . 00 

$748.00 

$261.00 

500 . 00 

$761.00 

$322.00 

453 00 

$775-oo 

$232.00 

453 00 

$685.00 

$302.00 
425-00 

$727-00 



$261.00 

389-00 

$650 . 00 

$232.00 

389 ■ 00 

$621 .00 

$232.00 
360.00 



25% 



$592 



$696 . 00 

948 . 00 

$1644.00 

$696 . 00 

948 . 00 

$1644.00 

$698 . 00 

850 . 00 

$1546.00 

$696 . 00 

794 00 

$1490.00 

$522.00 

794 • 00 

$1316.00 

$644 . 00 

723-00 

$1367.00 

$464 . 00 

723-00 

$1187.00 

$604.00 
679.00 



$1283.00 

$522.00 
628.00 



$1150.00 

$464 . 00 
628.00 



$1092.00 

$464 . 00 
584.00 



$1048.00 



148 



VESTANCES 

TABLE 71 — Continued 



Size, 


Type 


Loads 


H.p. 


100% 


75% 


50% 


25% 


18 


1 cy. horiz. 


Vd . 


$102. 50 


$139.00 


$205.00 


$410.00 






V a ... 


249 . 00 


272.OO 


336.00 


549.00 






v t ... 


$351-50 


$4II.OO 


$541.00 


$959 • 00 


25 


2 cy. vert. 


Vd . 


$102.50 


$139-00 


$205.00 


$410 . 00 






V a ... 

v t ... 


259.00 


282.OO 


349.00 

$5 54- 00 


571.00 




$361 . 50 


$42I.OO 


$981.00 


25 


1 cy. vert. 


Vd . 


$83.50 


$1 I I . OO 


$167.00 


$334 00 






Va... 

v t ... 


249.00 

$332.50 


272 .00 


336.00 
$503 . 00 


549.00 




$383-00 


$883.00 


32 


1 cy. horiz. 


Vd . 


$100.00 


$134.00 


$200.00 


$400 . 00 






V a . . . 

v t ... 


239.00 

$339-00 


260 . 00 


323-00 
$523-00 


529.00 




$394 . 00 


$929.00 


36 


2 cy. horiz. 


Vd . 


$116.00 


$i54-5o 


$232.00 


$464 . 00 






V a ... 

V t ... 


245 . 00 


265.50 
$420 . 00 


330.50 

$562.50 


544.00 




$361 .00 


$1008.00 


36 


3 cy. vert. 


Vd . 


$106.00 


$141-5° 


$212.00 


$424.00 






V a ... 

v t ... 


250.00 


271.50 
$413.00 


337-5o 
$549-5o 


555-oo 




$356.00 


$979.00 


40 


1 cy. horiz. 


Vd . 


$95-50 


$127.00 


$191 .00 


$382.00 






V a ... 


233.50 


250.50 


312.00 


51300 






v t ... 


$329-00 


$377-5o 


$5°3 • 00 


$895 . 00 


40 


1 cy. horiz. 


V D . 


$97-50 


$130.00 


$195.00 


$39° • 00 






V a ... 


223.50 


242.50 


302 . 00 


495-00 






Vt... 


$321.00 


$372.50 


$497.00 


$885.00 


50 


2 cy. horiz. 


Vd . 


— $97-50 


$130.00 


$195.00 


$390.00 






V a ... 


231.50 


251 .00 


322.00 


514.00 






v t ... 


$329.00 


$381.00 


$517.00 


$904 . 00 


5° 


1 cy. vert. 


Vd . 


$97-50 


$130.00 


$195.00 


$390 . 00 






Va.. 

v t ... 


223.50 


242 . 50 
$372.50 


302 . 00 

$497.00 


495.00 




$321.00 


$885.00 


So 


2 cy. vert. 


Vd . 


$83.50 


$111 .00 


$167.00 


$334 • 00 






V a .. 


231.50 


251 .00 


322.00 


514.00 






v t ... 


$315-00 


$362.00 


$489 . 00 


$848.00 


64 


2 cy. horiz. 


Vd . 


$97-50 


$130.50 


$i9S-5o 


$391.00 






V a . ■ 


223.50 


242.50 


301 . 00 


498 . 00 






v t ... 


$321.00 


$373-oo 


$496 - 50 


$889.00 



OIL ENGINES 

TABLE 71 — Concluded 



149 



Size, 
H.p. 



75 



75 



80 



80 



100 



Type 



1 cy. vert. 



3 cy. vert. 



2 cv. horiz. 



2 cy. horiz. 



1 cy. vert. 



2 cy. vert. 



4 cy. vert. 



Loads 



Vd $99 • 5o 

V a 210.50 

V t $310.00 

Vd $84.00 

V a 223.50 

V t $307-50 

V D $93-5o 

V a 2I7-OQ 

V t $310.50 

Vd $96.50 

V a 211 .OO 

v t $307-50 

Vd $96 • 50 

V a 203.00 

V t $299.50 

Vd $96.50 

V a 211 .OO 

Vt $307-50 

Vd $83 . 50 

V a 2I9-OQ 

Vt $302.50 



75% 



$133 


OO 


228 


OO 


$361 


OO 


$112 


00 


242 


00 



$354 • 00 

$134.00 

231-50 

$365 • 5o 

$128.50 

224.00 

$352-5o 

$128.50 
215.00 

$343 • 50 

$128.50 

224.00 

$352-5o 

$111 .00 
232.00 

$343 • 00 



SO c / 





$199 
284 


00 
00 


$483 


00 


$168 
301 


00 

00 


$469 


00 


$186 
292 


50 
5o 


$479 


00 


$193 
283 


00 
5o 


s$476 


50 


$193 

273 


00 
00 


$466 


00 


$193 
283 


00 
50 


$476 


50 


$167 
294 


00 

5o 


$461 


5o 



25% 



$398 
469 


00 
00 


$867 
$336 

499 


00 

00 
00 



$835.00 

$373 00 
482.00 

$855-00 

$386 . 00 
470.00 

$856.00 



$386 


00 


453 


00 


$839 


00 


$386 


00 


47o 


00 


$856 


00 


$334 


00 


490 


00 



$824 . 00 



ISO 
$1700 

1500 

1300 

o 
1 1100 

I 900 



VESTANCES 



700 



500 



600, 



\ 








































\ 








































\ 








































































































































































































1 






















2$l.#_I oad 














I 








































\ 








































X 


s 








































V 
























fiO#- Xoa 


1 










1 


\ 


^ v 




















10 


[*Load 


75 


Load 







10 20 30 40 50 60 70 80 90 100 

HJP. 

Fig. 26. — Total vestances per h.p. of horizontal oil engines. 



An inspection of the curve of total vestance at full load 
shows that the prices of the 36 h.p. engines are considerably 
too high, while that of the 2 cylinder 25 h.p. engine is also 
high. With these exceptions, the prices are remarkably con- 
sistent, with the prices of the vertical engines running con- 
tinually below those of the horizontal engines. Down to 30 
h.p. the increase in total vestance is not great, showing good 
adaptability for small isolated plants. 

103. Diesel Engines. — For comparison with other prime 
movers we give below the vestances of Diesel engines based 
on a normal year of 3000 hours, interest at 5 %, fuel oil at 0.05 
cent per h.p.h., and life of the engines 15 years. Attend- 
ance costs are included. 



DIESEL ENGINES 



151 



TABLE 72 

Vestances per H.p. of Diesel Engines 
4 cycle 



Size, 
H.p. 



60 



90 



no 



150 



300 



500 



750 



Loads 



100% 



Vd $148.50 

Va I37-SO 



V t $286.00 

Vd $129.50 

V a 129 .50 



V t $259.00 

Vd $123.00 

V a 126.OO 



Vt $249.00 

Vd $116.00 

V a I2I.OO 



Vt $237.00 

Vd $111.00 

V a II7-OQ 

V t $228.00 

Vd $106.00 

V a II3-OQ 

V t $219.00 

Vd $99- 50 

V a IO7.5O 



V t $207.00 

Vd $96 . 00 

V a IO5.OO 

V t $201.00 

Vd $93 00 

V a 103.00 



Vt ..- $196.00 



75% 



$198.00 

149.OO 

$347.00 




$164.00 

I36.OO 

$300 . OO 

$155.00 

I3O.OO 

$285.00 



$148.00 

125.OO 

$2 73-00 

$141 .OO 

I20.00 

$26l.OO 



$13200 

II5-OQ 

$247.00 

$128.00 
I I I . OO 

$239.00 

$124.00 

108.00 

$232.00 



$297.00 

175-00 

$472.00 

$259.00 
164.00 

$423 • 00 

$246.00 

158.00 

$404 . OO 

$232.00 

152.00 

$384.00 

$222.00 
146.00 

$368.00 

$212.00 

140 . 00 

$352.00 

$199.00 

1 34 00 

$333 00 

$192.00 

130.00 

$322.00 

$186.00 
127.00 

$31300 



25% 



$594.00 
246 . 00 

$840 . OO 

$5l8.00 

224.OO 

$742.00 




$464 . OO 

209 . OO 

$673.00 

$444 . OO 

202 . 00 

$646 . 00 

$424 . 00 
193-00 

$617.00 

$398.00 
185.00 

$583.00 
$384-00 

1 79 . 00 
$563.00 

$372.00 

I75-QO 

$547 ■ 00 



152 



VESTANCES 



$yuu 










































800 


















































































.700 

X 
1 600 




















































































































































25 


ELoad 










03 










































> 

"3 




V 








































\ 


s 


















































































V 


^ B 






















50 


# Load 






















































75 * Load 






































100^ Lcai 














200 




10 





20 





30 





40 





50 





600 


700 


800 


900 


1000 



H. P. 

Fig. 27. — Total vestances per h.p. of Diesel engines at fractional loads. 

104. Induction Motors. — Since the efficiency of induction 
motors is independent of the speed, the only difference in the 
vestance of induction motors of the same size and type will 
be in the depreciation vestance. Since the depreciation vest- 
ance is but a small part of the total vestance, there will be 
but little difference in the cost of service even though the 
first cost of low speed motors may appear to be very large in 
comparison with the first cost of high speed motors. We 
illustrate this below. 

Example 31. — Compare the total vestance of a 10 h.p. 
1800 r.p.m. 3 phase 60 cycle 220 volt induction motor with 
another motor of the same size and type but of only 600 r.p.m. 
speed. Assume the normal year at 3000 hours, interest 5 %, 
power cost one cent per h.p.h. and neglect other operating costs 
besides those of power. 

Solution: The first cost of the 1800 r.p.m. motor is $160 
or $16 per h.p. With 25 years of life and interest at 5%, the 
term factor is 0.70469, so that the depreciation vestance per 
h.p. is 

16 -7- 0.70469 = $22.75. 



INDUCTION MOTORS 153 

The efficiency of this sized motor is 87.9% so that the 
power used per h.p. year is 

3000 -T- 0.879 = 34 2 ° h.p-h.f 
costing 3420 X 0.01 = $34.20. 

The operating vestance is then 

34.20 -*- 0.05 = $684, 
and the total vestance is 

22.75 + 684 = $706.75. 

For the 600 r.p.m. motor the first cost is $300 or $30 per 
h.p. so that the depreciation vestance per h.p. is 
30 -=- .70469 = $42.50, 

while the operating vestance is the same as for the 1800 r.p.m. 
motor or $684. 

The total vestance is then 

42.50 + 684 = $726.50. 
The difference is 

42.50 - 22.75 = $19.75 
out of a total of $706.65 or 

19.75 -*- 7o 6 -75 = 2.66%. 

The only object in using low speed motors is for the pur- 
pose of combining them with other low speed apparatus and 
where this cost of combination may be reduced by an amount 
equal to or greater than the additional cost of the lower speed 
motor. We have chosen above an extreme difference in speed. 
Where the difference in speed is not so great, the difference in 
total vestance will be even less than this small amount of 
2.66%. 

105. We give below the vestances of three phase, 60 cycle, 
1800 r.p.m. induction motors based on a normal year of 
3000 hours, power cost one cent per h.p.h. and interest at 5 %, 



*54 



VESTANCES 



TABLE 73 

Vestances per H.p. of Induction Motors 

3 PHASE, 6o CYCLES, l8oo R.P.M. AT PULL LOAD 



Size 


Vestances 


Ratio 


H.p. 


Depreciation 


Operating 


Total 


Va 


I 


$56.80 


$767-00 


$823 . 80 


13-5 


2 


35-55 


728.OO 


763 


55 


20 


5 


3 


26.95 


713.OO 


739 


95 


26 


5 


5 


19.80 


700.00 


719 


80 


35 


4 


75 


25.60 


689 . OO 


7i4 


60 


26 


9 


IO 


22-75 


684 . OO 


706 


75 


30 





IS 


18.20 


677.OO 


695 


20 


37 


2 


20 


16.70 


674.00 


690 


70 


40 


3 


25 


14 -5° 


671 .OO 


685 


50 


46 


3 


30 


14-45 


669.OO 


683 


45 


46 


3 


40 


12.43 


667.00 


670 


43 


53 


6 


50 


n-37 


665 . OO 


676 


37 


58 


5 


75 


9.76 


663 . OO 


672 


76 


68 






Note the very great ratio of operating to depreciation 
vestance, and the very great total vestance. This we shall 
discuss further. 

Two phase motors usually run one to two per cent lower 
in efficiency, increasing the total vestance between $7 and 
$14 per h.p., a very appreciable amount. The operating 
vestance is very large though the power rate assumed is far 
lower than that which can ordinarily be obtained. The ten- 
horse power rate is usually 2 cents per h.p.h. On this basis 
the operating vestance would be doubled, increasing the ratio 
to 60. 

106. At fractional loads, the splendid full load efficiency of 
induction motors is well maintained, so that there is no great 
variation of vestances at fractional loads. This is shown in 
the table below. 



INDUCTION MOTORS 



155 



TABLE 74 
Vestances per H.p. of Induction Motors 

3 PHASE. 6o CYCLES, l8oO R.P.M. 



Loads 



125% 



Vd 

V a . 



$45 
791 



Vt $836.40 

Vd $28.40 

Va 750.00 



Vt $778.40 

Vd $25-50 

V a 73<3 OO 

V t $761.50 

Vd $15-85 

V a 722.00 

Vt.. $737.85 

Vd $20.50 

Va 7II.OO 

V t $731-50 

V D $18.20 

V a 705.00 



Vt $72320 

Vd $14-57 

V a 697.50 



V t $712.07 

V D $13-37 

V a 695.00 



V t $708.37 

Vd $11.60 

V a 692.00 



Vt $703.60 

Vd $11-55 

V a 69O.OO 



V t $701.55 



IOO% 



$56 . 80 
767.OO 



$823.80 

$35-55 
728.00 



$763-55 




$19.80 

700 . 00 

$719.80 

$25.60 

689 . 00 

$714.60 

$22.75 
684 . 00 

$706.75 

$18.20 

677.00 

$695 . 20 

$16.70 

674.00 

$690.70 

$14.50 

671.00 

$685.50 




75% 



$75 
767. 


75 
00 


$842 


75 


$47 
728 


30 
00 


$775 


30 


$42 
7i3 


50 
00 


$755 


50 


$26 
700 


40 
00 


$726 


40 


$34 
689 


10 
00 


$723 


10 


$30 
684 


30 
00 


$7i4 


30 


$24 

677 


30 

00 


$701 


30 


$22 
674 


25 
00 


$696 


25 


$19 
671 


30 
00 


$690 


30 


$19 

669 


27 
00 


$688 


27 



50% 



$113.60 

791 .00 



$904 . 60 

$71.10 

7 50.00 

$821 .10 

$53 90 

73600 

$789-90 

$39- 60 

722.00 

$761.60 



$51.20 
711 .00 

$762.20 



$45 ■ 50 

705-0° 

$750.50 

$36.40 

697-50 

$733-9o 




$29.00 

692.00 

$721.00 

$28.90 

690 . 00 

$718.90 



25% 

$227.20 

852-50 

$1079.70 

$142.20 

809.00 

$951.20 

$107.80 

792.00 

$899 . 80 



$79.20 
778.00 

$857-20 



$102.40 

766 . 00 

$868.40 

$91 .00 

760.00 

$851.00 

$72.80 

75200 

$824.80 

$66 . 80 
749 00 

$815.80 



$58.00 

746.00 

$804.00 



$57- 80 
744-QQ 
&801 . 80 



156 



Size, 
H.p. 



40 



50 



75 



VESTANCES 

TABLE 74— Continued. 



Loads 



125% 
V D $9 . 98 

V a 688.OO 

V t $697.98 

Vd $9.10 

V a 685.00 

Vt $694.10 

Vd $7-8o 

V a 684.00 

V t $691.80 



100% 



$12.43 
667.00 

$679-43 

$ii.37 
665 . 00 

$676.37 

$9- 76 
663 . 00 

$672.76 



75% 



$16.60 

667.OO 

$683 . 60 

$15-15 
665 . OO 

$680.15 

$13-00 

663 . OO 
$676.00 



50% 



$24.86 
688.OO 



$712 

$22 
685 



$707 

$19 
684 



$703 



52 



25% 



$49.72 
741 .OO 



$790 

$45 
739 



$784 

$39 
737 



$776 



72 



$1000 



900 



700 



GOO 



































































































































































1 
























































































































A 








































\ 


N 






































I 


















^25-tf 




rt 


















l 








































\ 
















Z 


50 f 


Loac 




^75 J 


' Lo; 


id 
























M25 


*L 


>ad 




SJ 


K T 



























































10 



20 



30 



40 



50 
H. P. 



60 



70 



Fig. 28. — Total vestances per h.p. of induction motors at fractional loads. 

There is comparatively little change of total vestance at 
fractional loads because the efficiency is so well maintained. 
This is a very valuable feature for maintaining good economy 
under actual load conditions, such as are invariably met in 
practice. 

107. Generators. — The vestances for alternating current 
generators of the direct connected type are given below based 



GENERATORS 



157 



on a normal year of 3000 hours, cost of energy \ cent per kv.a., 
interest 5% and a life of 25 years. 

TABLE 75 

Vestances per Kv.a. of D-c. Generators 

3 phase, 60 cycles, 220-550 volts 



Size, 
Kv.a. 



25 



50 



75 



150 



250 



500 



75o 



Loads 



Vd $28.40 

V a 208.IO 



V t $236.50 

Vd $22.70 

V a 182.OO 

Vt $204.70 

Vd $18.80 

V a 172.OO 

Vt $190.80 

$17.00 

169.50 



Vd 
V a . 

V t . 



.$186.50 



Vd $15-90 

V a 167.25 



Vt $183.15 

Vd $i4 N 8o 

Va 165.40 



Vt $180.20 

Vd $13.00 

V a 164.OO 



V t $177.00 

Vd $11.70 

V a l62. 25 



Vt $173-95 

Vd $10.85 

V a l6l .OO 



V t $171.85 

V D $10.67 

V a 160.50 

Vt $171-17 



100% 



$35 
205 


50 
00 


$240 


50 


$28 
178 


40 
00 


$206 


40 


$23 
169 


50 
00 


$192 


50 


$21 
166 


30 
80 


$188 


10 


$19 
i54 


90 
10 


$184 


00 


$18 
162 


50 
75 


$181 


25 


$16. 
161 


30 

25 


$177. 


55 


$14 
I59_ 
$173- 


20 

i°. 

70 


$13- 
158. 


65 

50 


$172. 


15 


$13- 
158. 


35 
00 


$171 


35 



75% 



$47 
209 


30 
80 


$257 


10 


$37 
182 


80 
50 


$220 


30 


$31 
172 


30 
50 


$203 


80 


$28 
170 


35 

25 


$198 


60 


$26 
168 


5° 
00 


$194 


50 


$24 
166 


70 

75 


$191 


45 


$21 
165 


70 

00 


$186 


70 


$18 
162 


90 

75 


$181. 


65 


$18. 
162. 


20 
25 


$180. 


45 


$17- 
161. 


So 
75 


$i79- 


55 



50% 



$71 

222 


.00 
.80 


$293 


.80 


$56 
193 


.80 
.00 


$249 


.80 


$47 
184 


.00 
.00 


$231 


00 


$42 
181 


60 
00 


$223 


60 


$39 
178 


80 

75 


$218 


55 


$37 
177 


00 
50 


$214 


50 


$32 
175 


60 
00 


$207 


60 


$28 
173 


40 
25 


$201 


65 


$27 
172 


3° 

50 


$199 


So 


$26 
172 


7o 

00 


$198 


70 



$142.00 

257.00 
$399.00 

$11360 

217.00 

$330.60 

$94 • 00 
205 . 00 

$299.00 



$85 . 20 
202.50 

$287.70 

$79 60 

200 . 00 

$279.60 

$74 • 00 

197-50 

$271.50 

$65 . 20 
I 95-5Q 

$260.70 



$56.80 

193-25 

$250.05 

$54 60 

192 50 

$247.10 

$53 • 4o 

I9I-75 

$245-15 



158 



VESTANCES 



$1200 



1100 



1000 



900 



800 



700 



600 



















































































































































































































































1 \ 








































\ 


s 






































1 






^ 


































IV 
























.25 Load 






































I 


^ 


«*»^ _ 




































A 


^ 




































s 


— 


















.75 Load 






























Ful 


1 Los 


d^ 




M.25 Full Load 

1 1 1 











100 200 300 400 500 600 700 800 900 1000 

"K. V. A. 

Fig. 29. — Total vestances per kv.a. of electric generators at fractional loads. 

Like motors, there is comparatively little variation in total 
vestance with the size of the unit or the load. This is one of 
the immensely valuable characteristics of electrical machinery 
that it is so well adapted to meet the load variations of prac- 
tice with remarkably good economy. 

108. Comparison of Power Units. — By power units we 
mean, primarily, such units as steam engines, turbines, inter- 
nal combustion motors, and electric motors. In the fore- 
going, we have determined the vestances of these units based 
on a normal year of 3000 hours, constant load, and 5 % interest. 
The fuel costs assumed were coal at $2 per ton with a fuel 
value of 13,743 B.t.u. or 5.4 h.p.h. per lb. Allowing for boiler 
service, we assumed a cost of $0,005 P er h.p.h. of energy de- 
livered to the engine. In comparison with this we assumed 
for oil engines, oil at 0.5 11 cent per pound, and allowing 320 
lbs. per barrel, this would amount to $1,635 P er barrel, or 
0.07 cent per h.p.h. For Diesel engines we assumed the 
cost of the heat energy at 0.05 cent per h.p.h. corresponding 
to a cost of $1,168 per barrel. And finally for motors we 
assumed a cost of one cent per h.p.h. At 15 cents per gallon 



COMPARISON OF POWER UNITS 159 

for gasoline, energy would cost 0.2675 cent per h.p.h. For 
clearness we tabulate these as follows: 



TABLE 76 

Cost of Energy per H.p.h. Assumed 

Cost of Energy per 
Unit H.p.h. in Cents 

Induction motor 1 . 0000 

Gasoline engine o . 2675 

Oil engine o . 0700 

Diesel engine o . 05 

Steam engine o . 05 

With these assumed costs clearly in view, for they vary 
greatly with different localities, we found comparisons as 
follows : 

At full load, the Diesel engine has the best economy, with 
the oil engine second, the compound condensing steam engine 
third, the simple low speed noncondensing steam engine 
fourth, the simple high speed noncondensing steam engine 
fifth, the electric motor sixth and the gasolene engine seventh. 

Below 30 h.p. the electric motor is far superior in economy 
to the simple high speed steam engine, and no doubt the other 
types of steam engines as well, but it is itself far inferior to the 
oil engine, somewhat inferior to the distillate engine at 7 J 
cents per gallon for the distillate, and far superior to the gaso- 
line engine. The sharp bending of the curve (Figure 30) of 
the Diesel to the horizontal for small sizes would lead one to 
expect the oil engine to be superior to the Diesel below 
40 h.p. 

109. We give below in tabular form the total vestances of 
these units for both full and fractional loads, together with 
the relative valuance based on unity for the machine which 
has the best economy, so as to show clearly their compara- 
tive values. Where the values are not given in the tables 
they have been interpolated from the curves. In the table, 
V t is the total vestance and U, the comparative value. 



160 VESTANCES 

TABLE 77 
Total Vestances and Comparative Value of Power Units 



Size, 
H.p. 



250 



5o 



25 



Unit 



Diesel engine 

Comp. steam engine. . 
S. L. S. steam engine 
S. H. S. steam engine. 

Diesel engine 

Oil engine 

Comp. steam engine. . 
S. L. S. steam engine. 
S. H. S. steam engine. 

Diesel engine 

Oil engine , 

S. H. S. steam engine 
Electric motor 



Oil engine 

Gasoline engine 

S. H. S. steam engine. 
Electric motor 



Loads Pek Cent of Full Loads 



307 
402 
448 



345 
442 

499 



556 
694 



095 
704 



1. 00 
0.76 
0.69 



1 .00 
0.78 
0.69 



1. 00 
0.80 



1 .00 
0.99 



Vt U 



223 
300 
386 
426 

254 
300 

339 
429 

474 

300 
321 
53o 
676 

332 

864 
660 
686 



1 .00 
0.74 
0.58 
0.52 



1. 00 
0.94 

o.57 
0.44 

1. 00 
0.38 
0.50 
0.48 



Vt U 



267 
315 
396 
434 

306 
345 
361 

443 
485 

365 
373 
545 
680 

383 
960 
682 
690 



r .00 
0.85 
0.67 
0.62 



0.89 
0.85 
0.69 
0.63 

1. 00 

0.98 
0.67 
o.54 

1. 00 
0.40 
0.56 
o-55 



V U 



360 

353 
440 

474 

414 
466 
412 
496 

532 

5io 

497 
603 
708 

503 
1212 

757 
721 



0.98 
1 .00 

0.80 
°-75 

0.99 
0.90 
1 .00 
0.83 
0.77 

0.98 

1 .00 
0.82 
0.70 

1. 00 
0.42 
0.66 
0.70 



Vt u 



631 

•457 
543 
560 

728 
839 
544 
609 

637 

910 

885 
732 
785 

883 

2089 

936 

804 



0.72 
1 .00 
0.84 
0.82 

0.76 
0.66 
1 .00 
0.91 
0.87 

0.80 
0.83 
1 .00 
°-93 

0.91 
0.38 
0.86 
1. 00 



$900 



800 





G* 


Bolin 


) Enj 


rine 






























































































































































Elec 


trie ] 


ilotor 






































1 
























































































































\ 


























s. 


fT.R. 




onde 


osin| 






\ 


























s, 


L.S. 


ion t 


onde 


asing 








\ 












v_ 
























"~ 












Oil E 


ngin 


B 










c 


)mpo 


and ] 


;.S. S 


team 


































1-En; 










• 













































. 700 



600 



400 



300 



200 



25 50 75 100 125 150 L75 200 225 250 
H. P. 

Fig. 30. — Total vestances of power units at full load. 



COMPARISON OF POWER UNITS 161 

It is evident from the above table that at full load the 
Diesel engine leads in economy by a wide margin, with the 
oil engine second and the compound steam engine third. For 
sizes over 50 h.p., the oil engine is worth only 85% as much 
as a Diesel and the compound steam engine about 75%. In 
other words, taking the cost of service of the Diesel as unity, 
then that of the oil engine is 1.15, i.e., 15%, greater and the 
cost of service with compound condensing engines is 1.33 or 
33% greater. 

At 75% load the Diesel still leads but at 50 h.p. the oil 
engine has practically become equal to it. Diesel engines 
are seldom built below this size, though a few have been made 
as small as 8 h.p. It is certainly remarkable how well the 
designers of engines have chosen the inferior limit of the 
sizes of their various types. 

At 50% load, the Diesel and compound engines have 
reached equality in sizes of 100 h.p. and over, with the oil 
engine third. But below 50 h.p. the oil engine is best. 

At 25% load, the compound steam engine leads by a wide 
margin above 100 h.p. At 50 h.p. the simple high speed steam 
engine is best, with the electric motor second, while below 25 
h.p. the electric motor leads with the oil engine second, based, 
however, on costs of energy as given. 

In all this the steam engines have the peculiar advantage 
that they can carry with comparative good economy 25% 
overload. The range of loads of these engines is therefore 
25% greater than internal combustion motors. The advan- 
tages of this greater range of load will be shown later. 

The gasoline engine is inferior to the electric motor at all 
loads on the basis chosen. But if, for example, 2 cents per 
h.p. is paid for electric power as compared to 15 cents per 
gallon for gasoline, then conditions would be reversed. 

In the face of the great superiority of the Diesel engines, it 
seems remarkable that they are not more generally used. 
There are two reasons for this. First the general resistance 
of engineers to anything new to them, and second the fact 
that there are many cases where the undertakings are under- 



l62 



VESTANCES 



financed, so that they are unable to put in the most economical 
installations. Above all this looms the element of uncer- 
tainty, so that there is a lack of incentive to build for 
permanency. 

110. Centrifugal Pumps. — In the case of centrifugal 
pumps, the costs of attendance, oil, and the like, are very 
small, as compared with the cost of power. In the following 
tables of vestances of centrifugal pumps, we have, therefore, 
left these items out of consideration. The vestances are given 
not only at full and fractional loads, but at variable heads as 
well. We have assumed a cost of power at one cent delivered 
at the pump shaft, interest rate 5% as before, and a normal 
year of operation at 1500 hours, with 25 years as the life of 
the pumps. 

TABLE 78 

Vestances per iooo G.p.m. of Capacity op Low Head Belted Centrifugal 

Pumps 



Rated 

Capacity, 

G.p.m. 


Deprecia- 
tion Yes- 
tance per 


Operating Vestances at Heads of 










IOOO G.P.M. 


10' 


20' 


30' 


50' 


50 


$936.60 


$3400 . 00 


$6000 . 00 


$7500.00 


$11,360.00 


100 


596.10 


2680.00 


4410.00 


5920.00 


9,360.00 


150 


482.50 


2350.00 


4050 . 00 


5370.00 


8,320.00 


225 


37750 


2025.00 


3625.00 


4670.00 


7,430.00 


300 


326.40 


1875.00 


3330.00 


4420 . 00 


7,070.00 


400 


265.40 


1790.00 


3120.00 


4170.00 


6,690.00 


700 


202 .80 


1565.00 


2770.00 


3750.00 


6,040.00 


900 


189.20 


1445.00 


2540.00 


3570.00 


5,850.00 


1,200 


177.20 


1390.00 


2500.00 


3460 . 00 


5,680.00 


1,600 


160.00 


1320.00 


2380 . 00 


336o.oo 


5,500.00 


3,000 


119.50 


1295.00 


2320.00 


3310.00 


5,460.00 


4oOO 


106.40 


1270.00 


2270.00 


3260.00 


5 ,400 . 00 


6,000 


96.70 


1250.00 


2230.00 


3220.00 


5,340.00 


7,000 


93 40 


1230.00 


2220.00 


3170.00 


5,310.00 


8,000 


91 . 20 


1210.00 


2200.00 


3120.00 


5,280.00 


10,000 


87.30 

83.40 


1190.00 
1185.00 
1170.00 


2 1 80 . 00 


3080 . 00 
3040 . 00 
3000 . 00 




14,000 


2170.00 






30,000 
40,000 
50,000 


78.00 
76.80 

76.20 

75-70 


2140.00 
2135.00 
2130.00 
2125 .00 










H55-00 
1150.00 

















CENTRIFUGAL PUMPS 



163 



$7000 


\ 








































6000 


\ 










































v 


— — 














































































5000 




















































































24000 

■s 


\ 








































> 


\ 








































|3000 


\) 


v. 












-20-ft 




d 
























\ 








































2000 


\ 


s_^ 












-10-ft 


-He« 


<\ 
































































1000 





















































































2.000 4000 6000 8000 10000 120.00 1400D 16000 18000 20000 
Capacity in G.E.M. 

Fig. 31. — Total vestances of low head belted horizontal centrifugal 
pumps at full load. 



iplOVVV 




\ 






































14000 




\ 














































































12000 




\ 








































\ 






































§10000 

02 




\ 










































\ 


















Wi 


;h75 


'Fra 


ne 












> 

3 8000 






> 


V 




































V 










































\ 




















Wit 


ii50' 


Frai 


ie 


























V 


N 


































4000 
































































w 


th25 


'Fra 


me 












2000 











































200 



400 600 
Fig. 32. — Total vestances of vertical centrifugal pumps 



800 1000 1200 1400 1600 1800 
Capacity in G.P.M. 



164 



VESTANCES 



TABLE 79 

Vestances per iooo G.p.m. or Capacity of Vertical Centrifugal 

Pumps 



Capacity, 


Vestances at Heads of 


G.p.m. 


?5 r 


50' 


75' 


100 
IS© 


Va $52IO.OO 

Vd 2780.00 

Vt $7990.00 

V a $4690.00 

Vd 2090.00 

Vt $6780.00 

V a $4150.00 

Vd 1570.00 


$9,370.00 

3,760.00 

$13,130.00 

$8,330.00 
2,770.00 


$13,700.00 

4,740.00 

$18,440.00 

$12,300.00 
3,440.00 


225 


$11,100.00 

$7,500.00 
2,100.00 


$15,740.00 

$11,050.00 
2,670.00 


300 


Vt $5720.00 

V a $3980.00 

Vd 1330 oo 


$9,600.00 

$7,070.00 
1,740.00 


$13,720.00 

$10,420.00 
2,140.00 


400 


Vt $5310.00 

Va $3750.00 

Vd 1063.00 


$8,810.00 

$6,690.00 
1 ,405 . 00 


$12,560.00 

$9,890.00 
1,710.00 


700 


Vt $4813.00 

V a $3290.00 

Vd 717.00 


$8,095.00 

$6,040.00 

938 . 00 

$6,978.00 

$5,860.00 
866.00 


$11,600.00 

$8,930.00 
1,140.00 


900 


Vt $4007.00 

V a $3020.00 

Vd 670.00 

V t $3690.00 

V a $2925.00 

Vd 587-00 

V t $3512.00 

V a $288o.OO 

Vd 378.00 

Vt $3258.00 


$10,070.00 

$8,650.00 
1 ,040 . 00 


1200 


$6,726.00 

$5,680.00 
727.00 


$9,690.00 

$8,410.00 
866 . 00 


1600 


$6,407.00 

$5,510.00 

657.00 

$6,167.00 


$9,276.00 

$8,160.00 

762.00 

$8,922.00 



CENTRIFUGAL PUMPS 



165 



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o a 

< 
ft 

< 
U 









•o ft 

c34 



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8 8 






O O 
O O 






















O O 
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oc 
















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to 
to 


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CENTRIFUGAL PUMPS 



167 



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VESTANCES 



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. 80Q JJ5O0 240.0 3200 4000 4800 5600 6400 7200 8000 

Capacity in G.P.M. 

Fig. S3- — Total vestances of horizontal direct connected centrifugal pumps. 

111. Nothing could show better how vestance indicates di- 
rectly the relative values of various equipment Than table 8o. 
Where there are so many different sizes, speeds, capacities, etc., 
each with a different efficiency and price, it would seem at 
first that there are so many variables that a basis of compari- 
son could not be found. When, however, all these variables are 
reduced to the basis of a permanent rendition of the service, the 
problem simplifies at once and we have the total vestance. 
That unit which, under given conditions, has the lowest total 
vestance is the most economical, i.e., has the highest financial 
efficiency. 

Thus at 25 ft. head, the 20 h.p. 2300 g.p.m. pump has the 
lowest total vestance of $2919 per 1000 g.p.m. of capacity, 
the next lowest being the 30 h.p. 3200 g.p.m. pump, having 
a total vestance of $3042. Each of these pumps has a first 
cost of $500, apparently, a rather high price. Yet the first cost 
represents in this case only about one-sixth of the total cost 
of rendering the service. 

In the case of the 20 h.p. 2300 g.p.m. pump, the first cost 
per 1000 g.p.m. of capacity is $218 and the total depreciation 



CENTRIFUGAL PUMPS 169 

vestance is $309. The difference of $91 is due to limited life. 
If then this pump could be so constructed that its life would 
be permanent, our total vestance would be reduced only by this 
amount of $91, not a very significant per cent of the total 
vestance. It is evident then that increasing the life of 
pumps is not nearly as important as increasing the pump 
efficiency. 

On the other hand, since the operating vestance is $2610, a 
saving of 10 per cent of the power would reduce our total 
vestance $261, a very significant amount. The question 
naturally arises as to how much the first cost could be in- 
creased under the above conditions for a saving of ten per 
cent of the power used. This is obtained by multiplying the 
saving in total vestance by the term factor which is 0.70469 
in this case. This gives 

261 X 0.70469 = $183.92. 

If, then, a pump of 2300 g.p.m. requiring 20 h.p. is worth $500, 
then a pump of the same capacity requiring 10 per cent less 
power or 18.0 h.p. will be worth 

500 + 183.92 = $683.92. 

If, for example, the cost of the latter were $600, it would still 
save the customer $83.92 and this is for each 10 per cent of the 
power saved. Such improvement by refinement in design and 
workmanship and by the use of better materials is attainable 
to the extent of some 5 % on the average pump built. 

It should be noted that according to the above list at 25 
ft. head, it would be more economical to use five 20 h.p. 2300 
g.p.m. pumps than two 50 h.p. 4800 g.p.m. pumps on a 100 h.p. 
installation. In the former case the pump cost will amount to 

5 X 500 = $2500, 
and in the latter case to 

2 X 675 = $1350. 

The saving in use of the 5-20 h.p. pumps in total vestance 
over that of two 50 h.p. pumps as above, will be 

$3324 - 2919 = $405 



i7o VESTANCES 

for each iooo g.p.m. of capacity or a total of $4657.50, cer- 
tainly a significant sum. 

At 50 ft. head the 40 h.p. 2600 g.p.m. pump is best, the 
next best being the 35 h.p. 200 g.p.m. pump, the difference in 
total vestance being 

$5558 - 5135 = $423- 

At 100 ft. head the 250 h.p. 7000 g.p.m. pump is best 
with the 200 h.p. 5600 g.p.m. pump a close second. 

Finally at 150 ft. head, the 75 h.p. 1400 g.p.m. pump is best 
with the 100 h.p. 1800 g.p.m. pump as a close second. 

112. It will be noted that the largest sized pump does not 
always represent the best investment considering the way pumps 
are now being manufactured. However, if it is possible at 50 
ft. head to make a 40 h.p. pump with a total vestance of only 
$5135 it certainly is possible for the manufacturer to make a 
150 h.p. pump with an even lower total vestance instead »of 
$830 higher per 1000 g.p.m. of capacity. It is apparent that 
the design of the latter needs revision. In the future when 
a new design has been made and tested, it will have to be 
valuated, the results of which will determine whether it will be 
adopted or junked. 

In the list of pumps at 25 foot head are two 10 h.p. pumps, 
one of which costs $240 and delivers 820 g.p.m., while the 
other costs $385 and delivers 1000 g.p.m. The difference in 
costs is $145 and the difference in capacity 180 g.p.m. Usually 
the latter pump is considered to be so expensive that its use 
is prohibitive except in "fancy" installations. Yet the differ- 
ence in vestance of $520 or nearly 15% shows that in reality 
the use of the cheap pump is prohibitive. 

At high heads, good efficiency is still more important than 
at low heads. At 150 ft. head there are listed two 100 h.p. 
pumps, one having a capacity of 1500 g.p.m. and the other a 
capacity of 1800 g.p.m. The latter has a first cost of $925, 
and the former a cost of $550, a difference of $375. That is 
the latter costs 68% more than the former. It is often said 
that such fine pumps as the latter are installed merely for 



CENTRIFUGAL PUMPS 171 

show. Yet the difference in total vestance in favor of the 
latter is 

20,520 - 17,380 = $3140 
or 3140 ■*■ 20,520 = 15.3%. 

This means that in the latter case the service can be rendered 
15.3% cheaper. 

It should be borne in mind that the cost of power assumed 
is very low, much lower than can usually be obtained. The 
effect of increased power cost is to make an increase of effi- 
ciency more valuable, and to increase the difference in vestances 
of units whose efficiencies differ. 

113. It is ordinarily assumed that pumps are either used 
at full capacity, or not at all. This is true only to 
the extent that pumps are not throttled. But almost in- 
variably the total head pumped against varies with the season, 
i.e., with the change in level between the supply water and the 
point of discharge. The point of discharge is usually constant, 
but the level of the supply water almost invariably changes. 
In the case where the water is pumped from a river, the water 
level may change as much as 15 or 20 ft. The effect of this 
is that if the head is decreased below normal, the capacity of 
the pump is increased above normal, i.e., the pump is over- 
loaded, while if the head is increased above normal, the pump 
will be underloaded. The amount of load variation is very 
small in high head pumping plants and usually large in very 
low head plants. 

It is evident, then, that the effect of fractional loading can- 
not be neglected since the efficiency of pumps falls off quite 
rapidly with either increase or decrease of capacity, especially 
the former. Unless we take into consideration the amount of 
water to be pumped at each stage of water level, the cost of 
service will be inexplicably increased. And it must be borne 
in mind that a very few feet variation in head will change 
the pump capacity very greatly, this being especially true of 
increased heads above normal. An idea of the effect of all 
this may be obtained from the following example. 



172 VESTANCES 

Example 32. — A centrifugal pump delivers 1000 g.p-m. 
against 66 ft. head at 60% efficiency, this being its point of 
best efficiency. Assuming a normal irrigation year of 1500 
hours, cost of power at 1 cent per h.p.h. and interest at 5%, 
and neglecting other operating costs, determine the operating 
vestance at full, three-quarters, half, and one-fourth loads. 

Solution: The power used at full load is 

66 X 1000 , 

— = 27.5 h.p., 

4000 X 0.60 ' ° * 

costing 27I cents per hour or 

0.275 X 1500 = $412.5 per year. 
The operating vestance is then 

412.5 -J- 0.05 = $8250. 
At three-fourths, or one and one-fourth load, this would be 

8250 -*- 0.937 = $8804.70, 

since at that load the pump only gives 93.7% of full load 
efficiency. 

So at half or one and one-half loads, the operating vestance 
is 8250 -f- 0.75 = $11,000, 

and at one-fourth load it is 

8250 -s- 0.436 = $18,920. 

114. It is evident that at less than three-fourths or over one 
and one-fourth load, the operating vestance becomes prohibi- 
tive, unless the pump is used but a few hours per year. The 
effect of the number of hours of use per year and the location 
of the change point is similar to that of steam engines, except 
that in this case we have assumed the normal year at 1500, 
instead of 3000, hours. 

Thus if Vq is the operating vestance of pumps per normal 
year as listed in the preceding tables, then the operating vest- 
ance per hour is — — and the total vestance per year of ( N) 
1500 

hours is 

V a N 



V D + 



1500 



CENTRIFUGAL PUMPS 173 

Equating the total vestances of any two pumps will deter- 
mine the change point. 

Example 33. — At 150 ft. head, two 75 h.p. direct connected 
pumps are listed, one having a capacity of 1200 g.p.m. and 
costing $500 and the other having a capacity of 1400 g.p.m. 
and costing $900. Determine the change point on the basis 
previously assumed. 

Solution: The operating vestance of the 1200 g.p.m. pump 
per hour is 

18,750 -5- 1500 = $12.50. 

The total vestance per year of (A) hours is 

Vi = 592 + 12.5 N. 
For the 1400 g.p.m. pump, the operating vestance per hour 
is 16,100 -r 1500 = $10,733, 

so the total vestance per year of (N) hours is 

V 2 = 915 + i°-733^- 
The change point is determined by equating the total vest- 
ances, i.e. 

592 + 12.5 iVi = 915 + 10.733 N l5 
or Ni = 182.8 hrs. 

If the pumps are to be used less than 182.8 hours per year 
the cheaper pump is best; for more than this number of hours 
of use the more efficient pump is best. 

The difference in total vestances of two units operating at 
different efficiencies increases with the number of hours of 
use. Thus when A" = 5000 hours, 

Vi = 592 + 12.5 X 5000 = $63,092, 
and V 2 = 915 + io-733 X 5000 = $54,580, 

giving a difference of 

63,092 - 54,580 = $8512. 
The difference at 1500 hours was only 

19,342 - 17,015 = $2327. 

It is evident then that for very few hours of use per year 
a cheap pump should be used. But as the number of hours 
of use per year increases not only should we use the more 



174 VESTANCES 

expensive pump, but the comparative value of the more effi- 
cient unit continually increases. What we pay for first cost 
is usually only a very insignificant part of the cost of service. 
The user, under normal conditions, can afford to pay much 
more in first cost to the manufacturer than is now commonly 
done for more efficient units. 

115. A considerable decrease in the life of units does not 
greatly increase the cost of service. Consequently increased 
efficiency should be obtained even if it is necessary to greatly 
decrease the life of the unit. Yet too often the reputation of 
a certain make of apparatus is based on the length of service, 
rather than on the quality and cost. On the other hand, there 
is a tendency to continue to use units when they have become 
inefficient due to wear or other causes, so that the service is 
no longer rendered economically. Under such conditions, the 
pumps should either be brought back to full efficiency, or else 
discarded for new apparatus. Because under such conditions, 
the excess operating vestance has become so great that the 
continued use of the worn equipment would be far more 
expensive than the cost of replacement. 

116. The vestances of such items of a system as belts, 
pipe, and the like may easily be determined, as illustrated 
below. 

Example 34. — A belt for a 10 h.p. unit costs $20 or $2 per 
h.p. The belt efficiency is 85%. Assuming power at one 
cent per h.p.h., interest at 5%, and the life of the belt at 5 
years, and neglecting other operating costs, determine the 
vestance per h.p. of the belt per normal year of 3000 hours. 
How does it compare with the vestance of a flexible, steel link 
coupling costing $45? 

Solution: With interest at 5% and the life at 5 years, the 
term factor is 0.21649. The depreciation vestance per h.p. is 
then 

2 -s- 0.21649 = $9.25. 

For each h.p.h. transmitted 0.15 h.p.h. is lost. The loss per 
h.p. year is then 



PIPE 175 

3000 X 0.15 = 450 h.p.h., 
costing 450 X 0.01 = $4.50. 

The operating vestance is then 

4.50 + 0.05 = $90, 

and the total vestance is 

9.25 + 90 = $99-25- 

If a 10 h.p. flexible coupling costs $45, then the cost per h.p. 

is $4-5°- 

For all practical purposes, the life of the coupling is per- 
manent. The depreciation vestance per h.p. is therefore 

4.50 -T- I = $4.50. 

The efficiency is 100%. The operating vestance, neglecting 
attendance and repair as we did for the belt, is then zero and 
the total vestance is 

4.50 + o = $4.50 

as compared with $99.25 for the belt. On this basis, it ap- 
pears therefore that we could spend as much as $94.75 per 
h.p. for direct connection and still render the service as cheap 
as the belt-driven unit in cost. 

If the power costs 2 cents per h.p.h., the operating vestance 
of the belt would be doubled ($180) and its total vestance per 
h.p. would increase to $189.25. We could then afford to spend 
this greater amount per h.p. for direct connection. 

As a matter of fact the difference between the belt and 
coupling is much greater than is indicated above. For, 
while the cost of attendance and repair, that we neglected 
to take into consideration above, is very small for the coupling, 
it is very large for the belt. Furthermore the belt places heavy 
lateral strains on the bearings of both the driver and driven 
unit. The loss of power and increased repair on the units 
that this occasions must be charged against the belt. In a 
well-made coupling, these losses are eliminated, the power 
being transmitted without such strains. 

117. Pipe. — Neglecting attendance and repair which is 
inappreciable in a well-designed installation, the cost of opera- 



176 



VESTANCES 



tion of pipe is the cost of the power lost in transmitting the 
fluid, usually water, in friction. This loss varies with the 
velocity at which the fluid is transmitted, increasing rapidly 
with increased velocity. 

The life of pipe has been the subject of much argument. 
If improperly used, the life may be very short, due to corro- 
sion. If, however, the pipe is covered with some noncon- 
ductor, such as asphaltum, or if it is taped and asphalted, this 
corrosion is eliminated. On the whole, it seems safe to assume 
a life of 50 years. 

We give below the vestances of pipe based on 5% interest, 
cost of power one cent per h.p.h. and also based on a normal 
year of 3000 hours. It must be borne in mind, however, that 
where the pipe is used for irrigation, the normal year should 

TABLE 81 

Vestances of Standard Iron Pipe per 100 Ft. 



Size, 

Inches 

Dia. 



Vd 

V a . 



Vd 

Va- 



Vd 
V a . 

V t . 



Vd 

V a . 

v t . 



Vd 

V a . 

v t . 

V D 

V a . 

V t . 



Velocity, Ft. per Second 



S3 -Si 
0.04 



$3-55 

$4.99 
0.05 



$5-04 

$8.23 

0.06 

$8.29 



$3 01 
o. 19 



$3- 20 
0.26 



$3-77 

$4.99 
o.39 



$5-38 

$8.23 

0.40 

$8.63 



$2 14 
o.54 



$2.68 



$2 

1 

$3 

$3 

1 

$4 

$3 

1 

$5 

$4 
2 

$7 

$8 

3 

$11 



$2.14 
11 .46 



$13 60 



$2.14 
15.40 



$17 

$3 

$22 

$3 

27 

$30 

$4 
3i 



$35 

$8 
44 

$52 



$2.14 
93.60 



$95 

$2 
106 



54 $108 



99 



$3 
144 



$147 

$3 
202 



$206 

$4 
232 



$237 

$8 

35i 

$359 



74 



$2 
660 


14 

.00 


$662 


14 


$2 
8lO 


14 
00 


$8l2 


14 


$3 
965 


01 
00 


$968 


01 


$3 
i,340 


Si 
00 


$i,343 


5i 


$4 
1,707. 


99 

75 


$1,712. 


74 


$8. 
2,300. 


23 
40 



$2,308.63 



PIPE 

TABLE 8 1 — Continued 



177 



Size, 

Inches 

Dia. 



2\ 



Velocity, Ft. pek Second 




$10,968.00 



be assumed at only half of this amount or 1500 hours. Cost 
of delivery, laying, etc., is not included. 

118. Galvanized pipe, running much higher in first cost 
than the black pipe, can be justified only by a proportionate 
increase in the life of the pipe, unless other advantages are 
forthcoming. This can hardly be done. Conducting cover- 



i 7 8 



VESTANCES 



$1400 



1200 



1000 



800 



$ 600 

o 

H 



400 



200 































| c 


3 


























loc 




























1 a 












/& 
















1° 
I®* 










A* 


f 
















Ij 








/ 


^ 


















r 






J 


A 








- 












1 






/ 






















' 




/ 




























/ 


























/ 
















r V; 


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> 


/ 


























/ 














^ 


»t.j 


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6 8 10 

Diameter in Inches 



12 



14 



Fig. 34. — Showing the variation of total vestance with the 
velocity, and size of pipe. 

ings have generally been displaced with nonconducting 
coverings. 

The above table shows that at very low velocity the operat- 
ing vestance is practically negligible in comparison with the 
depreciation vestance. For a 12 in. pipe, at 0.5 ft. per 
second of velocity, for example, the depreciation vestance is 
$168 while the operating vestance is only $0.30, or about 
0.18% of the total vestance. But at high velocities condi- 
tions are reversed. For example, at 20 ft. per second ve- 
locity, in the case of the 12 in. pipe, the operating vestance 
is $10,800 while the depreciation vestance is still $168. The 
latter, under these conditions, is only 1.53% of the total 
vestance and is thus almost negligible.- 

119. Somewhere between extremely low and extremely high 
velocity is the point of best economy. The above table does 
not show this. But this can be shown if the table is reduced 
to vestances per cubic feet per second (cu. sec. ft.) of capacity. 
This we have done in the table below. 



PIPE 



179 



TABLE 82 
Vestances per 100 Ft. of Standard Pipe per Cu. Sec. Ft. of Capacity 



Vd 

Va. 



Vd 

V a . 

v t . 



V D 
V a . 

V t . 



Vd 

Va. 
V t . 



V D 
V a . 

V t . 



V D 

Va. 

Vt. 



Vd 

V a . 
Vt. 



V D 

V a . 



Velocity, Feet per Secon 



$2265.00 
143.00 



Vd . . .$2340.00 

V a 28.00 

V t $2368.00 



$1870.00 
19.00 

$1889.00 



Vd . ■ .$1320.00 

V a 10.00 

Vt 



$1330.00 

$1045 . 00 
7.00 



.$1052.00 

$1035.00 
5.00 



$1040 . 00 



.00 
4.00 



$912.00 

$803 . 00 
3.00 



V t $806.00 



V D 

Va. 

V t . 



$734.00 
2.00 



$736.00 



$2408.00 

$1170.00 
87.00 



$1257.00 

$935- 00 
73.00 



$1008.00 

56o . 00 
32.00 



$3150.00 
795.00 



$1 260 . 00 
6750.00 



$3945 . 00 

$1430.00 
717.00 



$8010.00 

$570.00 
4110.00 



$2147.00 

$1132.00 
566 . 00 



$4680 . 00 

$453 • o° 
2890.00 



$1698.00 

$585.00 
311.00 



$3343-oo 

$234.00 
1820.00 



$896 . 00 

$468.00 
159.00 



$692.00 

$522.00 
26.00 



$548.00 

$518.00 
19.00 



$537- 00 

$454.00 
16.00 



$470.00 

$402.00 
15.00 



$417.00 

$367-00 
8.00 



$375- 00 



$627.00 

$330.00 
1 20 . 00 



$450 . 00 

$261.00 
96.00 



$357- 00 

$259.00 
71 .00 



$330.00 

$227.00 
54.00 



$281.00 

$201 .00 
38.00 



$239; 00 

$184.00 
29.00 



$213.00 



$2054.00 

$187.00 
1160.00 



$1347.00 

$132.00 
714.00 



$846.00 

$104.00 
566.00 



$670.00 

$103 . 00 
391.00 



$494 . 00 

$91 .00 
309 . 00 



$400 . 00 

$80.00 
211.00 



$291 .00 

$73 00 
167.00 



$240.00 



$630 . 00 

27,550.00 

$28,180.00 



$285.00 
14,140.00 



^4,425.00 

$227.00 
10,800.00 



$11,027.00 

$1 1 7 . 00 
7,500.00 



$7,617.00 

$94.00 
4,360.00 



$4,454.00 

$66.00 
2,810.00 



$2,876.00 

$52.00 
2,020.00 



$2,072.00 

$51.00 
1 ,490 . 00 



$1,541.00 

$45 • 00 
1,150.00 



$1,195.00 

$40 . 00 
816.00 



$856.00 

$37.00 

607 . 00 

$644 . 00 



$3i5-oo 
97,000.00 

$97,315.00 

$143.00 

54,000 . 00 

$54,143.00 

$113.00 

36,200.00 

$36,313.00 

$58.00 

22,350.00 

$22,408.00 

$47 • 00 
16,000.00 

$16,047.00 

$33 00 

9,200.00 

$9,233.00 



$26.00 

6,919.00 

$6,945.00 

$25.00 

4,670.00 

$4,695.00 

$23.00 

3,7i2.oo 

$3,735-oo 

$20.00 

2,635.00 

$2,655.00 

$18.00 

2,120.00 

$2,138.00 



i8o 



VESTANCES 

TABLE 82 — Continued 



Size, 
Inches 


Velocity, Feet per Second 


Dia. 


0.5 


1.0 


2.0 


5-0 


10. 


20.0 


6 


Vd . . . $632.00 

V a 2 . OO 


$318.00 
6.00 


$159.00 
22.00 


$63.00 
I22.00 


$32.00 

459.00 

$491.00 

$28.00 
355- 00 


$16.00 
1,586.00 


8 


V t $634.00 

Vd . • . $562.00 
V a 1. 00 


$324.00 

$28l.OO 
5.OO 


$l8l.OO 

$140.00 
17.OO 


$185.00 

$56.00 
92.OO 

$148.00 

$49 • °° 
71.00 


$1,602.00 

$14.00 
1,175.00 


10 


V t $563.00 

Vd . . . $494.00 

V a I . OO 


$286.00 

$247.00 
4.OO 


$157.00 

$148.00 
I3.OO 


$383.00 

$25.00 

262.00 

$287.00 

$22.00 
216.00 


$1,189.00 

$12.00 
963 . 00 


12 


Vt $495.00 

Vd . . . $448.00 

V a I.OO 


$251.00 

$2 24.00 
3.OO 


$l6l.OO 

$112.00 
IO.OO 


$120.00 

$45.00 
54.00 


$975.00 

$11.00 
721.00 




Vt $449.00 


$227.00 


$122.00 


$99 . 00 


$238.00 


$732.00 



bztsvv — 
























' 
















2400 -1 






















/ 




























/» 


:"pii 


e 






/ 


Vp 


pe 


























/ 










/ 




















1 








1 




































CD 

3 




/ 


































, 


SO 




/ 


l"P 


ipe 




























i9nn 


\ 




> 






f 






























_j 




5 




























^V' 


Pipe 








IV 






































1 


\\ 






































inn —5 


\v 
































"l2" 


Pipe 






V 















































































8 10 12 

Velocity in f.p.s, 



14 



16 



18 



20 



Fig. 35. 



Total vestances of standard iron pipe per 100 ft. of length 
and per cu. sec. ft. of capacity. 



We thus see at a glance that starting with low velocity the 
total vestance decreases as the velocity increases until it reaches 



PIPE 181 

a minimum, and thereafter increases. This minimum point is 
the point of best economy under the conditions assumed. If 
the cost of power is in excess of one cent per h.p.h., lower veloci- 
ties must be used, otherwise higher velocities for best economy. 
There are instances in the piping of water where some power 
is available but not sufficient for economical development. 
Under such conditions small pipe may be used, wasting this 
power in friction but reducing thereby the cost of the pipe. 
But there are other instances where the power is wasted merely 
because of a lack of knowledge of its worth. We have in mind 
one such instance where the power wasted was afterward put 
to use and found sufficient for a town of 3000 people. 

120. To give a better idea of the variation of total vest- 
ance per cu. sec. ft., with velocity, we give below more de- 
tailed data for the 6 inch pipe by way of illustration. 

TABLE 83 
Velocity — Ft. per Second 

0.5 0.6 0.8 1.0 1.5 2.0 3.0 4.0 5.0 6.0 8.0 10 15 20 



6" Fd. $632 $527 $395 $316 $211 $159 $105 $79 $63 $53 $39 $32 $21 $16 

V a - ■ 2 2 3 6 14 2 2 48 8l 122 iSo 298 459 IOOO 1586 

~ Vt. .$634 $529 $398 $322 $225 $181 $153 $160 $185 $233 $337 $491 $1021 $1602 

This shows that with the conditions assumed a velocity of 
three feet per second is the point of best economy while a 
velocity of four feet per second is nearly as good. At a velo- 
city of 20 ft. per second, the total vestance for the transmis- 
sion of water is over ten hundred per cent greater. 
• The friction of riveted pipe is considerably greater than 
smooth pipe, such as standard wrought iron or wood pipe. 
According to Mark's Mechanical Engineer's Handbook, that 
of 6 inch riveted pipe is 1.8 ft. per 100 ft. of pipe at 3 ft. per 
second velocity, while that of smooth pipe of the same size 
and at the same velocity is only 0.7 ft. under the same condi- 
tions; the operating vestance of the riveted pipe will be 
$122 as compared with only $47 for the smooth pipe. Since 
the total vestance of the latter is only $152, it would be neces- 
sary, for equal economy, to have only 
152 - 122 = $30 



182 



VESTANCES 



TABLE 84 

Vestaxces of Wood Pipe per 100 Feet per Cubic Second Feet at Three Feet 
per Second Velocity 



Size, 


Head m Feet 


In. . 

DlA. 


50 


100 


150 


200 


250 


300 


350 


400 


2 


Fd . . $178.00 

V a 203 . OO 

V t .. . .$381.00 

Vd . . $87.00 
V a .... nS-oo 


$182.00 
203 . 00 


$188.00 
203 . 00 


$203 . 00 
203 . 00 


$218.00 
203 . 00 


$232.00 
203.00 


$263.00 
203 . 00 


$299.00 
203.00 


3 


$385 ■ 00 

$91 .00 
115.00 


S391.00 

$93.00 
115.00 


$406 . 00 

$97.00 
115.00 


$421.00 

$101.00 
115.00 


$435-oo 

$108.00 
115.00 


$466 . 00 

$126.00 
115.00 


$502.00 

$138.00 
115.00 


4 


V t . ■ ■ .$202.00 

Vd • ■ $61.00 
V a .... 81.00 


$206 . 00 

$63.00 
81 .00 


$208 . 00 

$64.00 
81 .00 


$212.00 

$78.00 
81 .00 


$216.00 

$82.00 
81 .00 


$223.00 

$89 . 00 
81 .00 


$241 .00 

$98.00 
81 .00 


$253.00 

$108.00 
81.00 


6 


V t . . . .$142.00 

Vd . . $33 00 
V a .... 48.00 


$144.00 

$39.00 
48.00 


$145.00 

$42 .00 
48.00 


$15900 

$46.00 
48.00 


$163.00 

$50.00 
48.00 


$170.00 

$55.00 
48.00 


$179.00 

$59.00 
48.00 


$189.00 

$61. 00 
48.00 


8 


V t .. . . $81.00 
Vd . . $22.00 

V a . ... 34-00 


$87.00 

$28.00 
34.00 


$90.00 

$31.00 
34.00 


$94.00 

$35 00 
34.00 


$98 . 00 

$38.00 
34.00 


$103.00 

$41 . 00 
34.00 


$107.00 

$44 00 
34.00 


$109.00 

$51-00 
34.00 


10 


Ft.... $56.00 

Vd ■ ■ $20.00 
V a 26.00 


$62 .00 

$26.00 
26.00 


$65 . 00 

$28.00 
26.00 


$69.00 

$32.00 
26.00 


$72.00 

$35- 00 
26.00 


$75.00 

$38 . 00 
26.00 


$78.00 

$41 . 00 
26.00 


$85.00 

$47 00 
26.00 


12 


V t ... . $46.00 
Vd . . $1500 

V a 20 . OO 


$52.00 

$20.00 
20.00 


$54 00 

$23.00 

20.00 


$58.00 

$25.00 
20.00 


$61.00 

$29.00 
20.00 


$64 . 00 

$32.00 
20.00 


$67.00 

$37 00 
20.00 


$7300 

$41 . 00 
20.00 


16 


Vt.... $3500 

Vd ■ ■ $14.00 
V a 14.00 


$40 . 00 

$17.00 
14.00 


$43.00 


$45 00 


$49 ■ oc 


$52.00 


$5 7 . 00 


$61.00 
































V t .... $28.00 
Vd . 


$31.00 

$15.00 
11.00 




















V a .... 
















Vt 
















$26.00 

$11 .00 
9.00 




24 


V D . • $9.00 
V a 9 ■ 00 
































1 

1 










Vt... . $18.00 


$20.00 










1 









PIPE 



183 



$600 



Si 



500 



400 



O o 

V 0) 
on on 



8,300 



g * 200 

3 100 

o 
H 























































































































































































































































































































































































































































































































_20( 


ft. I 


ead 
















































"T" 





8 10 12 14 

Inches Diameter of Pipe 



16 



18 



20 



Fig. 36. — Total vestances of wood pipe at 50 and 200 ft. of pressure. 



of depreciation vestance. Assuming then that the riveted 
pipe has as long a life as the wrought iron pipe, i.e., 50 years, 
so that its term factor will be 0.91279, it would be necessary 
to purchase the riveted pipe at 

30 X 0.91279 = $27.38370 per 100 ft. 

as compared with $57.80 for the wrought iron pipe, i.e., the 
riveted pipe would be worth only 47.4% as much as the latter. 

The use of extra strong, or double extra strong, pipe is war- 
ranted where, and only where, the pressure to be borne is too 
great for standard pipe. For still lower pressures, casing may 
be used to considerable advantage- due to its lower first cost. 

121. We give below the vestances of wood pipe on the 
same basis as the standard pipe above. This pipe has come 
into very considerable use. It is particularly well adapted for 
low heads. It cannot, however, be allowed to stand empty 
without ruining it. The friction is assumed equal to that of 
standard pipe. 

Since the operating vestance of wood and standard pipe 
has been assumed equal, while the first cost, and therefore the 
depreciation vestance, of the former is much lower, the total 



184 VESTANCES 

vestance of wood pipe is far the lower. However, it must be 
remembered that there are some duties for which wood pipe 
is not adapted.' But in such service as the distribution of 
water for irrigation, wood pipe has proved an inestimable 
aid, allowing much development that would have been pro- 
hibitive with more expensive pipe. 

It would be a simple matter in any concrete case to com- 
pare the total vestances of pipe with that of open ditches or 
flumes. However, both the first cost and the cost of main- 
tenance of such conveyances is highly variable, varying greatly 
with the character of the ground, the conformation of the 
soil, and local weather conditions. In any concrete case, 
where these items are definitely known, exact comparisons 
can be readily made, so that we can determine which is 
the best. 

122. It must be borne in mind that, primarily the prices 
of all commodities are based upon the cost of production 
of these commodities. If a system is so installed that 
it does not render the service at best economy, then labor is 
being wasted. If the installation is too cheap in design and 
construction, then labor is wasted in operation. If, on the 
other hand, the installation is too expensive, then labor has 
been wasted in producing a system more refined and efficient 
than the amount and nature of the service warrants. Waste 
of brains, skill or labor is a dead loss; a waste that no com- 
munity, whether state or nation, can afford. In the face of 
international competition, such a waste leads with certainty to 
national impoverishment. It is a mark of incompetence. 

In the calculations of this chapter on motors, we have as- 
sumed a cost of i cent per h.p.h. This is, as we have pointed 
out, far lower than can be obtained in practice. In order to 
indicate what may actually be expected we publish the actual 
rates in Oregon for 191 9, as announced by the public service 
commission. 

These are as follows: 



PIPE 185 

TABLE 85 
Primary Rate 

First 500 kw.h. per mo. at 5 cents per kw.h. 

Next 500 " " " " 4 " " 
a 4;000 a a a a 3 a a 

" 10,000 " " " " 2 " " 
All excess " " " " i| " " 



Secondary Rate 

First 4,000 kw.h. per mo. at i| cents per kw.h. 
Next 100,000 " " " " 1.0" 
Excess kw.h. " " " 0.8 " " 



Assuming a load factor of 40%, i.e., assuming that the 
motors run at full load 40% of the time on the basis of a 24- 
hour day and are idle 60% of the time, and also allowing for 
the unavoidable motor losses, these rates are, for various sized 
motors, as follows: 

TABLE 86 

Corresponding mean rate 
Size of motors at 40% load factor 

Up to if h.p 5 cents per h.p. of motor 

if to 3 h.p 4% " " " " 

3 to 17.5 h.p 3.3 " " " " 

17.5 to 52.5 h.p 2.433 " " " " " 

52.5 to 105 h.p 2.004 " " " " " 

To determine the total vestance of a motor under such a 
rate, multiply the operating vestance by the rate for the sized 
motor under consideration and add thereto the depreciation 
vestance. 

For example in Table 74 we gave for a one h.p. motor at 
full load: 

Depreciation vestance $56 . 80 

Operating vestance 767 . 00 

Total vestance $823 . 80 

The operating vestance was based on 1 cent per h.p.h. Ac- 



1 86 VESTANCES 

cording to the above rate this should be 5 cents. For "this 

rate, the operating vestance would be 

767 x 5 = $3835, 

giving a total vestance of 

3835 + 56.80 = $3891.80 

instead of only $823.80. This would not compare well with 
engines of any type. 

We give in the table below the vestances of induction motors 
at full load with the rates as per the table above, other things 
as in the previous table on induction motors. 



TABLE 87 

Vestances per H.p. of Induction Motors at Full Load and 40% Load 

Factor 

as per the 19 1 9 Oregon rate . 







Vestances 




Size of Motop, 








H.p. 










Depreciation 


Operating 


Total 


1 


S56.80 


$3835-00 


$3891.80 


2 


35-55 


3276.00 


3311 


55 


3 


26.95 


2353-00 


2379 


95 


5 


19.80 


2310.00 


2329 


80 


ih 


25.60 


2273.70 


2298 


30 


10 


22.75 


2257. 20 


2279 


95 


15 


18.20 


2234.10 


2252 


30 


20 


16.70 


1640. 10 


1656 


80 


25 


14 -5° 


1632.80 


1647 


30 


30 


*4-45 


1627.90 


1642 


35 


40 


12.43 


1623.00 


1635 


43 


50 


n-37 


1618. 20 


1629 


57 


75 


9.76 


1326.00 


1335 


76 



CHAPTER VI 
UNIT COST DETERMINATION 

Unit Cost. Time Element. Change Points. Unit Cost at Constant 
Load. Nearly Constant Load. Two Part Load. Three Part Load. 
iV-part Load. Instantaneous Unit Cost of Actual Load Curve. 
Service Modulus and Replot. 

123. We have seen in the foregoing chapters that cost 
governs price, so that our real problem is primarily that of 
cost determination. But in what way shall we apportion the 
costs, when all elements that enter into the problem are vari- 
able? The answer to this is: That an equipment must earn its 
costs, and profit, while in operation — for certainly it cannot do 
so while idle. This is the fundamental basis of cost analysis. 

124. Let us then, to begin with, take the simplest condi- 
tions of operation possible, that in which the undertaking 
runs at full load for a variable number of hours per year. 

Let 

C = the cost per h.p. of the system, 
p = the per cent of fixed charges, 
a = the operating cost per h.p.h., 
L = the load carried and 
N = the number of hours of operation 

then, evidently 

pC = the fixed charges per h.p. year 
and LpC = the total fixed charges per year. 

The greatest value that (N) can have is 8760, the number 
of hours in a year. Again since the operating cost is assumed 
to be constant at (a) dollars per h.p.h. (or other unit of service), 
then La = the total cost of operation per hour, 

and LaN = the total cost of operation per year. 

187 



188 UNIT COST DETERMINATION 

Evidently then the total production cost (Ki) per year is 
the sum of the total annual cost of operation and the fixed 
charges or 

K x = LpC + LaN = L(pC + aN) . . . (69) 

If now we divide this total annual cost by the load (L) 
and the hours of operation (2V), we will get the cost per h.p.h. 

(K 2 )or r K x L(p C + aN) 

K2 'LN~ TN ' 

or K 2 = ^- + a (70) 

As (N) becomes less, the unit cost (K 2 ) becomes greater, 
until when N = o, the unit cost is infinite, the operating cost 
per h.p.h. being assumed constant. On the other hand as ( A 7 ) 
increases, the unit cost becomes less, tending to decrease to 
the value of the operating cost, when (N) = <» . But the 
largest value that ( A T ) can have is 8760, the number of hours 
in the year, and therefore the minimum value that the unit 
cost can reach is 

A 2 = — — + a 
8760 

when the plant is operated at full load throughout the entire 
year. 

125. If we have two plants, one of which has a first cost of 
Ci dollars, a depreciation of pi per cent, and an operating cost 
of d\ dollars, while the second has corresponding values, C 2 , 
p 2 , and (h and each carry the same load L, then the total 
annual cost of the first plant is 

AY = L{p 1 C 1 + a,N), 
and of the second is 

K{' = L(p 2 C 2 + a 2 N). 

If we plot the total annual production cost vertically as in 
Fig. 37 and the hours horizontally, then these equations will 
give straight lines AB and A ' B' '. It may be and often is true 
that two such lines as AB and A ' B' intersect between N = o 



THE CHANGE POINT 



189 

































































































































+J 




































^B 






c 



a 



















































































- 
-a 











































a 
c 




































-b' 
















C 






























< 
c 


a' 
















































































A 





























































































































Hours of Operation per Year 8760 

Fig. 37. — Variation of production cost with use and the change point. 

and N = 8760, as at (C), Fig. 37. At this change point 
Ki and Ki are equal and therefore 

or ^iCi + a x N = P2C2 + a 2 A^, 

and solving for N gives the change point 



N = 



- P 2 ^ 2 ~~ P 1 ^ 1 



(71) 



(ai — a 2 ) 

For values of {N) less than N , Ki is less than Ki.* For 
values of (N) greater than N , K/ is less than K{ . Evi- 
dently then when the hours of operation are less than N , the 
first plant is the most economical, for a greater number of 
hours of operation, the second plant is best. The application 
of this will be illustrated in the problems below. 

Example 35. — A pumping plant, having a capacity of ten 
thousand gallons per minute costs $8000 and has 14% fixed 
charges. The cost of operation of the plant is 20 cents per 
hour. What is the cost per hour per 1000 gallons of capacity? 

Solution: Under the above conditions the fixed charges are 
8000 X 14% = $1120, 



190 



UNIT COST DETERMINATION 



or $112 per 1000 gallons of capacity. The total cost of opera- 
tion is 20 cents per hour or 2 cents per 1000 gallons of capacity. 
For ( N) hours this cost would be o.02iV dollars. 

The total production cost per thousand gallons of capacity 
per year is then 

K x = 112 + 0.02N (a) 

The cost per hour is obtained by dividing (Ki) by the num- 
ber of hours (N), getting 



K, 



112 



-f- 0.02 



(b) 



We can get values for Ki and K 2 from equations (a) and (b) 
for various values of (N) as follows; 



TABLE 



N 


Ki 


K2 





$112.00 


infinite 


100 


114.00 


$1.14 


1000 


132.OO 


0.132 


3000 


172.OO 


0.0573 


5000 


2 I 2 . OO 


0.0424 


7000 


252 .OO 


. 0360 


8760 


287 . 20 


0.03278 



Plotting the values of K% vertically against the time (N) 
horizontally as in Fig. 38 gives a hyperbola. Note the enor- 
mous variation of the unit cost (K 2 ) with time. The number 
of hours of operation per year is the most important factor in 
the cost of a service. 

Example 36. — A power plant costs $100 per h.p. and is of 
2000 h.p. capacity, running at full load. If the fixed charges 
are 15% and the cost of operation 0.5 cent per h.p.h. (a) what 
is the total annual cost; and (b) what is the cost per h.p.h.? 

Solution: The total cost of the plant is 
2000 X 100 = $200,000, 
and the total fixed charges are 

200,000 X .15 = $30,000. 



CONSTANT LOAD 



191 



The operating cost per year is 

2000 X 0.005 X N = 10N . . . 
The total annual production cost per h.p. is then 



-^- = ^ + 0.005 

2000 A 7 N 



(a) 
(b) 



$1.40 



1.20 



ci.00 



.60 



o .40 
O 



.20 
























































































































































































































































<U 

a. 






































$30 







































3 






































9 


£ 










a^: 


t\oo < 


:osti 


,er^ 


eM 






















O 






to 


aVE 
























































































































Cost p 


er 1000 G.l 


3 .M. per Hour 
















1 


1 H — 









1000 2000 3000 



4000 5000 6000 
Hours per Year 



7000 8000 8760 



Fig. 38. — Variation of total and unit production cost with use, 
according to example 35. 

The values of K x and K 2 for various values of (N) are 
given below, in order again to illustrate the enormous varia- 
tion in cost with the time of operation. 

TABLE 89 



N 


Ki 


K2 





$30,000 . OO 


infinite 


100 


31,000.00 


o.i55 


1000 


40,000 . OO 


0.02 


3000 


60,000.00 


O.OI 


5000 


80,000 . OO 


0.008 


7000 


100,000.00 


0.007143 


8760 


117,600.00 


0.006712 



192 UNIT COST DETERMINATION 

Note that for iooo hours of operation, the cost per h.p.h. is 
30 times as great as for continuous ( N = 8760 hours) opera- 
tion during the year. 

Example 37. — A pumping plant of 1000 g.p.m. capacity 
costs $2400. The fixed charges are 12 % and the cost of opera- 
tion is 0.3 cent per thousand gallons delivered. Each acre 
irrigated requires 43,200 gallons of water per month, so that 
with 12 hours of operation per day, we can irrigate 50 acres. 
We wish, however, to irrigate 100 acres, and in as much as 
night irrigation is not feasible, the question is, shall we put in 
another pumping unit, or shall we build a reservoir, pumping 
into it during the night? If we put in the reservoir, the 
pumping plant will operate 24 hours per day and therefore 
the power company will reduce our rates so that the cost of 
the water will be reduced to 0.2 cent per 1000 gallons of 
capacity. The reservoir costs 25 cents per yard to excavate 
and the land costs $200 per acre. The reservoir permits of 
only 10 ft. depth so as to keep the bottom above the land 
level. The fixed charges on the reservoir are 8% and the 
length of the irrigation season is 5 months. Shall we install 
another pumping plant, or put in the reservoir? 

Solution: The question can only be answered by determin- 
ing the costs in each case and comparing them, the one ren- 
dering the service cheapest being the one we shall decide upon. 

Where we are operating 12 hours per day our costs are as 
follows : 

Fixed charges per year are 

2400 X 0.12 = $288, 
and the operating costs are 

$0,003 X 60 = $0.18 per hour, 

and since 5 months = 5X30X12 = 1800 hours the annual 
cost of operation is 

.18 X 1800 = $324. 

The total annual cost is then 

Ki = 288 + 324 = $612 per year. 



CONSTANT LOAD 193 

If we install two units, this cost will be doubled, or 
Ki = 2 X 612 = $1224. 

To hold the water pumped during 1 2 hours, the capacity of the 
reservoir will have to be 

1000 X 60 X 12 = 720,000 gal. 
or 720,000 ■¥ 7.5 = 96,000 cu. ft., 

or 96,000 -J- 27 = 3556 yards, 

costing 3556 X 0.25 = $889. 

If the reservoir is 10 ft. deep, it will cover 
96,000 -f 10 = 9600 sq. ft., 

which, allowing for some wastage of ground will use up about 
I acre, costing 

200 X i = $100, 

so that the total reservoir costs are 

889 + 100 = $989, 
on which the fixed charges are 

989 X 0.08 = $79.12 per year. 

With one pumping unit in combination with the reservoir, the 
total fixed charges per year are 

288 + 79.12 = $367.12 

and the operating costs will be, at the reduced power rates, 

0.002 X 60 = $0.12 per hour, 
or 0.12 X 3600 = $432, 

five months of operation amounting to 3600 hours now, based 
on 24 hours per day of operation. 
The total annual costs are then 

367.12 + 432 = $799.12. 

To summarize then, 

(a) With two pumping units running 12 hours per day for 
5 months, each of 1000 g.p.m. capacity, the total annual cost 
is $1224. 

(b) With one pumping unit of 1000 g.p.m. capacity running 
24 hours per day for 5 months, and thus pumping an equal 



i 9 4 UNIT COST DETERMINATION 

amount as the two units under (a) above, in combination with 
the reservoir, the total annual costs are only $799.12, showing 

a saving of 

1224 — 799.12 = $424.88 per year 

or 424.88 -v- 799.12 = 53%. 

126. Example 38. — A centrifugal pumping unit of 1000 
g.p.m. capacity costs $1100 and is guaranteed to operate at 
60% efficiency. A triplex plunger pumping unit of the same 
size costs $2400 and is guaranteed at 75% efficiency. If in 
either case the fixed charges are 15% and the power charges 
are 1 cent per h.p.h., other operating costs being 1 cent per 
hour, which equipment should we use, the head in either case 
being 80 ft.? 

Solution: (a) In the case of the centrifugal pump: 
The fixed charges are 1 100 X 0.15 = $165 per year, 

. , . 1000 X 80 , 

the power consumed is — = 33.3 n.p., 

4000 X 0.00 

The cost per hour for power is then 

33.3 X 0.01 = $0,333, 

and the total operating cost per hour is 

o-333+ °- 01 = $0,343, 

or °-343 N P er Y ear °f N hours. 

The total annual cost is therefore 

#1 = 165 +0.343N. ...... (a) 

(b) In the case of the triplex pump: 

the fixed charges are 2400 X .15 = $360, 

.i a • IOO ° X 80 , , 

the power consumed is = 26.7 n.p., 

v 4000 X 0.75 

so that the cost per hour for power is 

26.7 X 0.01 = $0,267, 

or 0.277 N per year of N hours. 

The total annual cost is therefore 

K x ' = 360 + 0.277N (b) 



CONSTANT LOAD 195 

For N = 1000 hours per year, 

Ki = 165 + 343 = $508 for the centrifugal pump, 
and Ki = 360 + 277 = $637 for the triplex pump, showing 
the centrifugal pump to advantage. 

For N = 5000 hours per year, 

then Ki = 165 + 1715 = $1880 for the centrifugal pump, 
and Ki = 360 -f 1385 = $1745 for the triplex pump, show- 
ing the advantage to rest with this pump. 

We can determine for what number of hours per year the 
two costs will be equal, by equating (a) and (b), thus 

165 + 0.343 Ni = 3°° + 0.277 Ni 
so that 0.066 JVi = 195 

or Ni = 2955 hrs. per year. 

That is, if we operate less than 2955 hours per year, then under 
the conditions assumed above, the centrifugal pump is best; 
if we operate more than 2955 hours per year, then the triplex 
pump is best. 

Example 39. — We wish to generate power under full load 
conditions and find that a hydro-electric plant will cost $100 
per h.p. and the cost of operation will be 0.1 cent per h.p.h- 
A Diesel engine plant will cost $60 per' h.p. and cost 0.2 cent 
per h.p.h. for operation, while a steam plant will cost $45 per 
h.p. and 0.3 cent per h.p.h. to operate. If the fixed charges 
in all cases above are 14% for what hours of operation would 
each of the above plants be best? 

Solution: The total annual fixed charges per h.p. for the 
hydro-electric plant are 

100 X 0.14 = $14. 
For the Diesel plant, they are 

60 X 0.14 = $8.40, 
and for the steam plant, they are 

45 X 0.14 = $6.30. 

The annual cost of operation for the hydro-electric plant per 
h.p. is $0,001 N. For the Diesel plant, it is $0,002 N, and for 



i 9 6 UNIT COST DETERMINATION 

the steam plant, it is $0,003^, where in each case N is the 
number of hours of operation per year. The total annual 
cost per h.p. for the hydro-electric plant is then 

K x = 14.00 + o.ooiTV^. 
For the Diesel plant, it is 

Ki = 8.40 + 0.002N, 
and for the steam plant, it is 

K^ = 6.30 + 0.003N. 
Equating K{ and 2Ti", we get 

6.30 + 0.003^1 = 8.40 + o.oo2iVi, 
so that o.ooii^i = 2.10, 

and iVi = 2100 hours per year. 

So also equating Ki and Ki, we get 

8.4O + 0.002iV 2 = I4.OO + O.OOliV2 

so that o.ooiiV 2 = 5.60 

or iV2 = 5600 hours per year. 

The conclusions then are: 

(a) If the plant is to operate less than 2100 hours per year, 
use the steam plant. 

(b) If the plant is to operate more than 2100 hours per year, 
but less than 5600 hours per year, use the Diesel plant. 

(c) If the plant is to operate more than 5600 hours per year, 
use the hydro-electric plant. 

127. If the plant does not operate at full load but at a con- 
stant fractional load, then calling M the capacity of the plant, 
and L the load, as before, we get that the annual fixed charges 
are MCp, the annual operating cost is LaN, the total annual 

cost is 

K x = MCp + LaN, 

and the cost per h.p.h. is 

K x MCp . e v 

K * = LN = -NT +a (72) 

and the application is as above. 



PROBLEMS 197 

Example 40. — A steam generating plant running at 80 % 
capacity costs $50 per h.p. If the fixed charges are 12% and 
the operating costs are 0.3 cent per h.p.h., what will be the 
annual cost per h.p. and the cost per h.p.h.? 

Solution: Here M = 1, L = 0.8, p = 0.12, a = $0,003 an d 

C = $50, 

whence Ki = 1 X 50 X 0.12 + 0.8 X 0.003^ 

or K\ = 6 + 0.0024^ (a) 

ii tz 1 X 50 X 12 . 

and also K 2 = J A7 h 0.003 

0.0 X iv 

or K 2 = 7 -^ + 0.003 (b) 

While in the above we have shown the application of the 
cost analysis, under special conditions, only to pumping and 
power generation, it is equally applicable to railroad or wagon 
road analysis as to dry goods or grocery business. 

PROBLEMS 

1. A pumping plant of 2000 g.p.m. capacity costs $2100. The fixed 
charges are 13%, and the cost of operation is 35 cents per hour. The plant 
operates for 2000 hours per year at full load. 

(a) What is the total annual cost of operation per year? 

(b) What is the total cost of operation per hour per thousand g.p.m. of 
capacity? 

2. If in problem (1), the plant operates for (N) hours per year: 

(a) What is the total cost of operation per year? 

(b) What is the total cost of operation per hour per thousand g.p.m. of 
capacity? 

(c) Plot the curves. 

3. An electric pumping unit of 10 h.p. size costs $240, while a gas engine 
driven unit costs $600 complete. The fixed charges in either case are 12%. 
The cost of operation of the electric pumping unit is 2 cents per h.p.h., 
while that of the gas engine driven unit is 1.2 cents per h.p.h. Determine 
for what number of hours of operation at full load each is best. Plot the 
curves. 

4. A 10 h.p. gas engine electric lighting unit costs $90 per h.p. and runs 
at full load for 3000 hours per year. The load increases during the period 
of operation to 15 h.p. We can take care of this additional load, either by 
installing another 5 h.p. unit as above, or by putting in a storage battery 



198 UNIT COST DETERMINATION 

and running the 10 h.p. unit 4500 hours per year. The battery costs $75 
per kw. of capacity and has an efficiency of 65%. The fixed charges on the 
above units are 15%. The operating cost of the engine unit is 1.3 cents 
per h.p.h. and of the battery 0.2 cent per kw.h. Should we install the 
extra engine unit or the storage battery for best economy? 

5. In problem (4), if the load increased to 20 h.p., which should we 
install? ■ 

6. In problem (4), if the load increased only to 12 h.p., which should we 
install for best economy? 

7. Given the following data: A horizontal 4-valve engine generating 
unit costs $22,700. The engine runs with steam at 150$ pressure, ioo° 
superheat and with a 26 in. vacuum, and uses 16. 5$ steam per kw.h. A 
steam turbine generating unit of the same size costs $12,250 and operates 
under the same conditions as the steam engine unit except with a 28 in. 
vacuum. It consumes 17.7$ steam per kw.h. The plants are run at full 
load. If the fixed charges on either plant are 14% and the steam costs 
18 cents per iooo#, and other operating costs are 0.06 cent per kw.h., for 
what length of service per year is each best adapted? 

8. A hydro-electric power plant costs $120 per kw.h. and operates at 
60% of full load, for (N) hours per year. Determine: 

(a) The total annual cost per kw. of the service; 

(b) The cost per kw.h. of the service; 

(c) Plot the curves for (a) and (b). 

9. We find that: a hydro-electric plant will cost $100 per kw. to install 
and c.2 cent per kw.h. to operate; a Diesel plant will cost $70 per h.p. 
to install and 0.3 cent to operate; while a steam plant will cost $50 per 
h.p. to install and 0.4 cent to operate. The fixed charges on the hydro- 
electric plant are 14%, on the Diesel plant 15% and 16% on the steam 
plant. ' The plants are to operate at full load. For what number of hours 
of operation is each of the above plants best? 

10. A pumping plant of 1000 g.p.m. capacity delivers water against 
120 ft. head with a guaranteed efficiency of 70% at full load and 40% effi- 
ciency at half load. The plant costs $850 to install. The power costs one 
cent per h.p.h., while fixed charges are 12%. The plant operates at full 
load for 2000 hours per year and at half load for 1000 hours per year. Deter- 
mine the total annual cost of operation. 

11. If in problem (10), the plant operated for (iVi) hours per year at full 
load and for (7V 2 ) hours per year at half load, what would be the total annual 
cost of operation? 

12. In problem (11), plot the curve for the variation of the total annual 
cost of production with (iVi) if N\ + iV 2 = 3000 hours. 

13. We are irrigating 100 acres of land for 5 months in the year with a 
25 h.p. direct connected electric centrifugal pumping unit costing $2200. 
The pump efficiency is 60%, motor efficiency 85%, while the head pumped 



PROBLEMS 



199 



against is 40 ft. total. The electric power costs 1.5 cents per kw.h. The 
plant runs at full load for 8 hours each day. We wish to irrigate 300 acres 
of land and can do so with the unit already installed by running 24 hours 
per day, in which case the power will cost us only one cent per kw.h. But 
in that case, we must install a reservoir to avoid night irrigation, costing 
35 cents per cubic yard to excavate and having a maximum depth of 12 ft. 
The land costs $100 per acre. Fixed charges on the pumping unit are 12%, 
and other operating costs are 3 cents per hour. The fixed charges on the 
reservoir are 10%, with an annual cost of $78 for maintenance and repair. 
With the reservoir in use the head on the pump will be increased by 4 ft., 
with a corresponding increase in power consumed. 

Instead of using the above layout, we can install two more 25 h.p. pump- 
ing units, the three units operating for 8 hours per day, or we can install 
one more such 25 h.p. pumping unit together with sufficient reservoir to 
avoid more than 8 hours of operation. In the case of the three 25 h.p. 
units, our power cost will be 1.3 cents per kw.h. and 1.2 cents for the two 
25 h.p. units. In the latter case the head will be increased by 2 ft., with 
a corresponding increase in power consumed. 

Determine which of the three possible types of installation are best, and 
by what per cent. 

14. A 10 h.p. direct connected centrifugal pump operates at 60% effi- 
ciency at full load, for 2000 hours per year. We wish to operate for another 
1000 hours per year with a load of 5 h.p. If we use the above pump, its 
efficiency will drop to 45%, while if we install a 5 h.p. unit it will give 
52% efficiency. The power costs 2 cents per h.p.h. while fixed charges are 
12%. Should we install the additional unit or use the 10 h.p. unit at half 
load? By what per cent would the one exceed the other? The current is 
a.c. 60 cy. 3 phase 220 volts. 

15. We wish to install a 10 h.p. electrically driven centrifugal pumping 
plant. Which would be best, (a) a belt driven unit or (b) a direct connected 
unit? Take into consideration cost of units, building and foundations. 
How much would we lose by installing the poorer one? 

Assume a power cost of 2 cents per kw.h., fixed charges 10% and 1500 
hours per year of operation at full load pumping against a 50 ft. head. 

16. Determine the same as for problem (15) but for a 100 h.p. plant 
with a power cost of 1 cent per kw.h., other things being equal. 

17. Determine the same as for problem (15) but for a one h.p. plant 
with a power cost of 3 cents per kw.h., other things being equal. 

18. Determine the same as for problem (15) but for a 50 h.p. plant with a 
power cost of 1.5 cents per kw.h., other things being equal. 

19. A 250 h.p. pumping plant is to operate against a head of 100 ft. 
Into how many units should we divide this plant for best economy, taking 
into account cost of building and foundations? Power cost one cent per 
kw.h., hours of operation 2000 per year and fixed charges 10%. 



200 



UNIT COST DETERMINATION 




20. Determine the same as in problem (19) for a 150 h.p. plant operating 
against 50 ft. head, other things being the same. 

21. Determine the same as in problem (20) but for a 30 h.p. plant operat- 
ing against 25 ft. head. 

128. Thus far we have considered an equipment operating at 
one uniform load during the entire year. If the load varies with 

the time, as shown by the 



curve, 00, Fig. 39, but not 
sufficiently so to necessitate 
the discontinuance of opera- 
tion of part of the plant, 
then for any given period 
we determine the mean load 
(L m ), and base our cost 
analysis thereon, instead of 
on the actual, variable load 
(M). In an electric gener- 
ating plant, the kw.h. produced, divided by the time in hours, 
gives the mean load. In general, the total production divided 
by the time in hours 
gives the mean load or 
average production per 
hour, i.e., the mean rate 
of production. 

But now, suppose we 
have a twin load as illus- 
trated in Fig. 40, i.e., a 
load of (Z2 + Z1) for (N 2 ) 
hours per year and follow- 
ing that, a load of (Z-i) 
for (iVi — N 2 ) hours per 
year. It is our problem 
then to determine the 

cost of production for a given plant, already in operation, or else 
the more difficult and important problem to so design the plant 
that the production will be at a minimum cost. In this case the 




-N— N5- 



Fig. 40. — Diagram for a two-part load. 



TWO-PART LOAD 201 

difference between L2 and L\ is assumed to be so great that it 
would be uneconomical to operate the entire plant for the 
lesser load, the entire plant operating for the load (Li -f- Z, 2 ) 

and ( T * T J per cent operating for the load L h 

The difficulty in the solution of this problem comes in the 
equitable distribution of fixed charges, primarily. But this 
difficulty is removed if you bear in mind that an equipment 
or unit thereof must earn its costs and profits while in opera- 
tion, for certainly it cannot do so when idle. Suppose then 
that this plant consists of two units, one of (Li) size and the 
other of (Z, 2 ) size. Evidently then the unit (Li) will operate 
for (A^i) hours per year, while the unit (Z 2 ) operates for only 
(2V 2 ) hours per year. During the balance of the year, above 
(Ni) hours, i.e., for iV hours (where 2Vi + N Q = 8760 hours) 
the load is assumed to be zero. 
For the unit (Li) we have 

LiCp = total fixed charges per year 
and LiaNx = total operating costs per year, 
where L\ = the load, 

C = cost of plant per unit of load, 
p = per cent of fixed charges, 
a = operating cost per hour per unit of load 
and A r i = number of hours of operation. 

Whence the total annual cost ( K u ) is 

K n = UCp + UaNx 
and the cost per unit of load (h.p.h., kw.h. etc.) per hour (K21) 

is K 2 i = — - + a. 

So, also, if only the unit (Z, 2 ) were in operation for the (A^ 2 ) 
hours, then the total annual cost K12 being 

K 12 ' = L 2 Cp + L 2 aN 2 
the cost per unit of load per hour is 

K 22 ' = ^ + a. 

1\2 



202 UNIT COST DETERMINATION 

The total cost for operating the unit (L 2 ) for (N 2 ) hours is 
K 12 ' = L 2 Cp + L 2 aN 2 

and for operating the unit (Li) for N 2 hours the total cost is 
obtained by multiplying the cost per hour (Eq. 68) by the load 
(Li) and the hours N 2 , i.e., by LiN 2 , getting 

kw = hMl + LiaN2 . 

The total cost of both units operating for N 2 hours is then 
the sum of Kn and K 12 " or 

K 12 = L 2 Cp + M^> + L 2 aN 2 + L,aN 2 , 
i.e., 

and the cost per unit of load per hour is obtained by dividing 
by the total load (£1 + i 2 ) and the hours N 2 or 

_ ( UN. + UNA Cp 
22 V NJft ){U+U) 

+ "K + li](iSz3 + «--<»» 

for the period (N 2 ), while for the period (A^i — N 2 ) it was, by 
equation (68) 

and nothing for the period (iV ) , since it is useless for this period 
according to the assumptions made in the problem. 

129. Example 40. — A steam power plant costs $50 per h.p. 
and is of 2000 h.p. size. It carries a load of 2000 h.p. for 3000 
hours of the year and 500 h.p. for an additional 2000 hours. 
If the cost of operation is 0.3 cent per h.p.h. and fixed charges 
are 12%, determine the cost per h.p.h. for each period, the 
load being considered integral. 

Solution: We can determine the costs by substitution in 
the equations, or work them out in detail as follows: 



TWO PART LOAD 203 

The 500 h.p. unit will operate for 

3000 + 2000 = 5000 hours. 

The annual operating cost of this unit is 
500 h.p. X $0,003 X 5000 hrs. = $7500, 

and the fixed charges are 

500 h.p. X $50 X 0.12 = $3000, 

so that the total annual cost is 

Ki = 3000 + 7500 = $10,500. 

The cost per h.p. year is 

10,500 -s- 500 = $21, 

and the cost per h.p.h. is 

$21 -r- 5000 = $0.0042. 

During the first 3000 hours we have an additional load of 
2000 — 500 = 1500 h.p. 

whose operating cost per year is 

1500 h.p. X 3000 hrs. X $0,003 = $13,500, 

and the fixed charges are 

1500 h.p. X $50 X 0.12 = $9000. 

The total cost of this part of the unit is 

K" = 9000 + 13,500 = $22,500. 

This is for the 1500 h.p. unit during the first 3000 hour period. 
But during this period we have in use the 500 h.p. unit as well, 
costing 

500 X 0.0042 X 3000 = 6300, 

so that the total cost during this 3000 hour period is 
22,500 + 6300 = $28,800, 

and since the load is 2000 h.p., the cost per h.p. is 
28,800 -5- 2000 = $14.40, 

and since this period is 3000 hrs., the cost per hour during this 
period is 14.40 -f- 3000 = $0.0048 per h.p.h. 



204 UNIT COST DETERMINATION 

Summary. — Cost per h.p.h. during ist period $0.0048. 
" " " " 2nd " 0.0042 

By substitution in the equations, we get 

rjr , 1 2 X 50 . „ 

A 2 = h 0.003 = 0.0012 -f 0.003 = $0.0042, 

3000 X 2000 

for the second period, and 

„ 2000 X 3000 -f- 2000 X 2000 — 500 X 2000 

A2 = 7 r 

2000 X 3000 (3000 + 2000) 

X 0.12 X 50 + 0.003, 
or 

K 2 = 0.0018 + 0.003 = $0.0048 per h.p.h. for the 

second period, as before. 

Example 41. — A factory for the manufacture of a gas en- 
gine costs $4,000,000 on which the fixed charges are 15%. 
The capacity of this plant is 600 engines per month. It sup- 
plies this number of engines per month during the first five 
months, and then 300 engines per month for the balance of 
the year, due to reduced demand. Exclusive of fixed charges, 
the cost of manufacture of each engine is $150. Determine the 
total cost per engine (a) during the first period, and (b) during 
the second period. 

Solution. — If we had a factory capacity of only 300 engines 
per month, its cost would be $2,000,000 on which the fixed 
charges would be 2,000,000 X 0.15 = $300,000 per year, or 
300,000 4- 12 = $25,000 per month; and since 300 engines 
are turned out per month, this gives 

2500 -f- 300 = $83.33 per engine, 

wherefore, for the slack period, the total engine cost is 

150 -f 83.33 = $233.33. 

The second $2,000,000 is required for the extra demand of 
300 engines per month during the first 5 months. The fixed 
charges thereon are 

2,000,000 X 0.15 = $300,000, 



TWO-PART LOAD 205 

which must be paid during this 5-month period, together with 
T 5 2 of the other $300,000 of fixed charges or 
T 5 2 X 300,000 = $125,000. 
Total fixed charges assignable to this first period are 
300,000 + 125,000 = $425,000 
or 425,000 -r 5 = $85,000 per month. 

During this period, 600 engines per month are manufac- 
tured, wherefore the fixed charges per engine during this rush 

period are 

85,000 -f- 600 = $183.33, 

and the total cost per engine during the maximum demand 
period is 

183-33 + !5o = $333-33, 
or $100 per engine more than during the normal demand period. 

If these engines are sold for $400 per engine, the net profit 
per month during the rush period is 

(400 - 333-33) X 600 = $40,000, 
and during the normal period it is 

(400 - 233.33) X 300 = $50,000. 

It is no doubt surprising that the net profits should be less 
during the rush period than during the normal demand period, 
but this is due to the large investment to accommodate this 
peak period. In this case it would be far better to reduce 
the size of the plant so that it could meet the total annual de- 
mand by producing engines at a uniform rate throughout the 
year. This would necessitate the tying up of capital on prod- 
uct finished ahead of the demand. The interest thereon 
must be charged against these off-season engines plus the cost 
of storage, but this would hardly amount to more than $5 
extra per engine as compared with $100 per engine extra on 
the first 3000 engines per year. 

130. The application to railway transportation problems is 
exactly similar as illustrated in the following problem: 

Example 42. — A branch railway 60 miles long costs 
$1,800,000 for ways and structures. Freight cars cost $3500 



206 UNIT COST DETERMINATION 

each and locomotives, one for each 20 cars, cost $30,000 each. 
Freight cars empty weigh 5 tons each and carry 30 tons net 
weight. Locomotives weigh 30 tons. Cost per ton-mile for 
freight haul is 0.4 cent. The freight load amounts to 120 
cars per day during July, August, and September, and to 20 
cars per day during the balance of the year. Besides this, 
passenger service is rendered. Passenger cars weigh 7 tons, 
costing $5000 each, and carry, on the average, 40 passengers. 
Cost per ton-mile of passenger service is 0.6 cent. The pas- 
senger load amounts to 4 cars per day during June, July, and 
August, and 2 cars per day during the balance of the year. 
Determine the total cost per ton-mile during rush and normal 
demand periods for passenger service, also the cost per pas- 
senger-mile if the fixed charges on ways and structures are 
10%, cars 14%, and locomotives 14%. 

Solution: The first problem is one of cost segregation as 
between passenger and freight service. The proper basis is 
evidently the ratio of the maximum demand. Thus the maxi- 
mum demand of passenger service consists of 

4 passenger cars @ 7 tons each 28 tons 

4 X 40 X i5o# = 24,000 (wt. of passengers) 12 " 

and 1 locomotive @ 30 tons 30 " 

Total weight . 70 " 

for 2 X 60 = 120 miles per day or 70 X 120 = 8400 ton-miles 
per day. 

For the freight service, the maximum demand consists of: 

120 cars per day @ 35 tons (one way) 4200 tons 

60 empties returned for loads assumed @ 5 tons (one 

way) 300 " 

6 locomotives @ 30 tons (two ways) 180- " 

so that (4200 + 300) X 60 = 270,000, 

180 X 120 = 21,600 
or 291,600 ton-miles per day. 

Of the $1,800,000 for ways and structures 

— ^— X $1,800,000 = $52,000 approximately 
291,600 



TWO-PART LOAD 207 

is to be charged to passenger service, and the balance, or 
1,800,000 — 52,000 = $1,748,000, to freight service. 

Of structures it should be understood that certain items must 
be charged to the particular service at once instead of being 
apportioned. Freight sheds and yards should be charged to 
freight service and passenger depots to passenger service 
directly, and so forth. 

The segregation of investments is then as follows: 

TABLE 90 
Passenger Ser\tce 

Cost F. C rate F. C total 

Ways and structures $52,000 10% $5,200 

4 passenger cars 20,000 14% 2,800 

1 locomotive 30,000 14% 4,200 

Total .. $12,200 

Freight Ser\ice 

Cost F. C rate F. C total 

Ways and structures $1,748,000 10% $174,800 

180 cars 630,000 13% 81,900 

5 locomotives 150,000 14% 21,000 

Total $277,700 

Segregation of Passenger Fixed Charge Costs — Normal Period 
Ways and structures S2600 

2 passenger cars 1400 

1 locomotive 4200 

Total m $8200 

Si 2,200 — 8200 = S4000. 
The normal load continues for 12 months, so that the fixed 
charges per month are 

8200 -=- 12 = S683.33. 

The excess load continues for 3 months during which the 
excess fixed charges are 

4000 ^-3 = S1333.33 per month. 
Total fixed charges per month during rush period are 

1333-33 + 683.33 = $2016.66 
or 2016.66 -v- 30 = $67.22 per day. 



208 UNIT COST DETERMINATION 

During normal period it was found to be only $683.33 P er 
month, or 

68 3-33 "*- 3° = $22.77 Per day. 

During the normal period, the train weight is 

Locomotive 30 tons 

2 cars 10 " 

80 passengers @ 150$ 6 " 

Total ^6 " 

or 46 X 120 = 5520 ten-miles per day. 

Whence the fixed charge per ton-mile is for the normal period 
22 -77 -*■ 552o = $0.0041. 
The total cost per ton-mile is 

0.0041 + 0.006 = $0.0101 during normal period. (Ans.) (a) 
Passenger miles per day during normal period are 

2 X 40 X 120 = 9600. 
Ton-miles per passenger-mile are 

5520 -v- 9600 = 0.575, 
whence the total cost per passenger-mile is 

0.0101 X 0.575 = $0.00581. (Ans.) (b) 

During the rush period, the train weight is 

Locomotive 30 tons 

4 cars 20 " 

160 passengers @ 150$ 12 " 

Total 62 " 

or 62 X 120 = 7440 ton-miles per day. 

Whence the fixed charge per ton-mile during the rush period is 
67.22 -*- 7440 = $0.00903, 
and the total cost per ton-mile is 

0.00903 + 0.006 = $0.01503. (Ans.) (c) 

Passenger miles per day during the normal period are 
4 X 40 X 120 = 19,200, 



VARIABLE LOAD — TWO PART 209 

so that the ton-miles per passenger-mile are 
7440 -*- 19,200 = 0.3875, 

and the total cost per passenger-mile during the rush period is 

0.01503 X 0.3875 = $0.005824. (Ans.) (d) 

In a similar way the freight costs per ton-mile during each 
period may be determined. Certain simplifications, hardly 
warranted in practice, have been made in the above problem 
in order to illustrate the application of the principles without 
too much detail. 

131. In former equations, it was assumed that the size of 
the unit and the load were equal, — i.e., the unit operated at 
full load. This is hardly ever the case. Suppose then 

(Li + L 2 ) = size of load during the rush period, 
{Mi + M 2 ) = " " units for the rush period, 

A^ 2 = time in hours (days or months) in the rush period, 
Li = size of load during the normal period, 
Mi = " " unit for the normal period, 
(Ni — N 2 ) = time in hours (days or months) in the normal 
period, 
C — cost per unit of load, 
p — fixed charges in per cent, 
and a = operating cost per unit of load per hour. 

Then evidently the total annual cost for the Mi unit is 

Kn = MiCp + UaNi (74) 

and the cost per unit of load is 

jt Mi Cp , , . 

K *=WiLi+ a (75) 

The total cost of the excess load during the rush period is 
K 21 ' = M 2 Cp + L 2 aN 2 . 

During this same period, we also operate the (Mi) unit whose 
cost for A^ 2 hours is 

All 



2io UNIT COST DETERMINATION 

whence the total cost for the rush period is 

Kl2 = (m 2 + ^A Cp + L 2 aN 2 + UaN 2 , 

or K n = (M 2 + M^A Cp + (L 2 + L x )aN 2 . . (76) 

and the cost per unit of load per hour (h.p.h., kw.h. or the 
like) is 

v (M x , M 2 \ Cp , ■ , . 

* 22 H^ + ^)rfr 2 +a (77) 

during the rush period as compared with 
_Jf t Cp 

for the normal period. 

132. Limitations on the value of (Ni) must of course be ob- 
served. If the time is in hours, then (iVi) cannot be greater than 
8760; if in days, then it cannot be greater than 365; and if in 
months, then (iVi) cannot be greater than 1 2 for evident reasons. 
In a great many cases it is hardly sufficient to divide the 
load into two parts. Electric service naturally divides itself 
into three distinct periods, the day, peak, and night periods. 
So the load on a great many undertakings naturally divides 
itself into three periods, the rush, normal, and slack periods. 
Under these conditions, we can determine the costs during each 
period as follows: 

Li = the mean load during the slack period, 
Mi = the capacity of unit during the slack period, 
Ni — N 2 = duration of the slack period, 
(Li + £2) = mean load during the normal period, 
(Mi + M 2 ) = capacity of units during the normal period, 
N 2 — Nz = duration of the normal period, 
(Li + hi + L 3 ) = mean load during the rush period, 
(Mi + M 2 + Ms) = capacity of units during the rush period, 
(N3) = duration of the rush period, 
C = unit cost of the undertaking, 
p = per cent of fixed charges, 
a = operating cost per unit of load per hour, 



THREE-PART LOAD 

then for the unit Afi, the total annual cost is 

K n = MiCp + LMh . . . 
and the cost per unit of load per hour is 

M x Cp 



211 



K- 



^ + a. 



(78) 
(79) 



This is for the slack period. 

L 



-Nr 



u 



N 
Fig. 41 . — Diagram for a three-part load. 

The excess load during the normal over the slack period is 
(L2) with a corresponding difference in size of unit of (M 2 ) } 
the total annual cost of which is 

K n ' = M 2 Cp + UaN 2 , 
and the cost per hour is 

KJ- = **£* + a. 

N 2 L 2 

During this period it operates for (N 2 — Nz) hours, for 
which time the cost is 



At 2 



Nz) + Lia{N 2 - N 3 ) 



to which must be added the cost of operating the (Mi) unit 
for the same period, or 

Kn'" = ^Cp(N 2 - Nz) + L ia (N 2 - Nz), 



212 UNIT COST DETERMINATION 

giving a total for this normal period of 

K l2 = 0£ + j£\Cp(N a - Nt) + (U + L 2 )a(N 2 - N s ) (80 



Dividing through by the load (L x + L 2 ) and the hours 
(N 2 — Nz) gives the cost per unit of load per hour, or 

_ (M, M 2 \ Cp ( , 

which is identical with equation (77). 

For the rush period the total cost of operating the excess 
unit (M 3 ) is 

K l3 ' = MzCp + LzaNz. 

The cost of operating the units (M 2 + Mi) for Nz hours is 
obtained by multiplying equation (77) by the load (L x + L 2 ) 
and the hours Nz, or 

K i3 " = 0£ + ~)CPN 3 + (L, + L 2 )aN 3 . 

The total cost of operation during the rush period is the 
sum of K 13 and K 13 ", or 



K B = \m 3 + (Ml + Ml\ N 3 ~\cp + (£, + U + U)aN 3 (8 



2) 



so that the cost per hour, ..obtained by dividing equation (82) 
by (£, + U + L 3 )N 3 is 

_/M 1 M t MA Cp , , 

K * ~ \T 1 + T 2 + T 3 )(L^n^+L 3 ) + a - (83) 

If we had (M) divisions in our load instead of only three as 
above, then the costs during the wth period can be written at 
once by symmetry, getting 

+ (!,! + Z*+. . >+Ln)Xa(N n -Nn + i) (84) 



N-PART LOAD 213 

total cost during the period, and therefore 

cp 

U + . • . + L„ 
(8s) 



* 2 " = U + ^ + ' ' ' + wjZT¥ 



133. We wish to call particular attention to the term 

(Ml Mi MA 

V#i # 2 ' NJ 

which we shall call the service modulus, and denote by (S), 
so that 

„ _ M 1 Mi 

62 _ Wt + ~Nl' 

'. Mi M2 Ms 

5s = ^ + ^ + Iv? etc " 

c Mi .M 2 . ,M n , . 

and s - = n; + ni + - ■ - + w n ■ ■ ■ (86) 

We shall also denote by L n ' the total load during any period, 
so that 

L n ' = Li -f- L2 + • • • + L n , 

whence we may rewrite equations (79) to (85) above as follows: 

#21 =Si^+ a, (87) 

Li 

Kn=Sz¥- + a, . . . ■ (88) 

K 2i = Sz^- + a, (89) 



and in general 



Cp 

Kin = S n j^ + a. . . (90) 



















K 


< 3 t 5 








• 


ii 


C ->2 


. 


j» 








t 

r. 


v 




c G 3 


' 



214 UNIT COST DETERMINATION 

134. If we plot the time (N) vertically and (S), the service 
modulus, horizontally as in Fig. 42, then the area is the size of 
the unit (M) required for the period. Thus 

« Mi 

Sl = Wi 

so that S1N1 = Mi, etc., the size unit required. 

In determining the costs of an undertaking, the work is 
greatly simplified by first determining the service modulus 

when the costs for each 
N period may be written 

down at once. 

Example 43. — A cer- 
tain plant carries a load 
as follows: first period, 
4000 hours per year, mean 
load 400 kw., maximum 
FlG 500 kw.; second period, 

2000 hours, mean load 
800 kw., maximum 1000 kw. ; third period, maximum 2000 
kw., mean load 2000 kw., time 2000 hrs. First cost per 
kw. $70, fixed charges 10%, and operating cost 35 cents per 
kw.h. Determine the cost per kw.h. during each period. 

Solution: In this case the service modulus (S3) for the third 
period, where 

Mi = 500, A r i = 8000, 

M 2 = 1000 — 500, A^ 2 = 4000, 

M s = 2000 — 1000, N3 = 2000, 

Z 3 = 2000 kw.h., 
is equal to 

c 500 . 1000 — 500 2000 — 1000 

03 = 1 1 , 

8000 4000 2000 

or S 3 = 0.0625 + °- I2 5 + °-5 = 0.6875. 

Whence the cost per kw.h. during this period is 

K 2S = 0.6875 x 70X0 ' 10 0.0035, 
2000 

K23 = 0.0024 + 0-0035 = $0.0059 per kw.h. (Ans.) 



SERVICE MODULUS 215 

Similarly the costs during the first and second periods may be 
determined. 

135. In general, we had for the service modulus, 

■ *~Ai + ;V 2 + ' ' ' + iVn-i N n 
for in) periods. 

For the in + i)st period, it is 

Mi , M 2 , , Af n -i , M n , M n+1 



n+1 



+ ^ + ' ' • + tF^ + F 1 + 



#1 ^2 A r n-1 #» #»+! 

the difference between the two being 



M n+1 



N n+1 

When the number of periods is indefinitely increased, then 
this difference becomes indefinitely small and we may write 

S n +1 — S n = dS. 

At the same time M n +i becomes very small, so that we may 

write 

M n+1 = dM, 

whence M = — (91) 



and therefore 



ff +» M 



when B = a constant of integration. 

The determination of the constant of integration may offer 
some difficulty. Since the limits of integration are from 8760 
hours to o hours, we must use one of these limits in determin- 
ing this constant. This must be the former. For we do not 
know the value of (S) when N = o, but when N = 8760, then 

5 = Si = — * = r—^, where (Mi) is the load when N = 8760. 
Ni 8760 ' 

Mi 
So then B = 7—7-, and our formula becomes 
8760 



/ 



'dM , M 1 . . 

If + Wo (93) 



2l6 



UNIT COST DETERMINATION 



The integration is really carried out from right to left instead 
of the usual order. 

At the same time the difference between the maximum de- 
mand (M) during a given period and the mean load vanishes, 
so that 

L = M 
and the cost per unit of load per hour becomes 

K 2 = S^+a (94) 

136. If we plot the hours of operation horizontally as in 
Fig. 44A and the load (M) vertically, then we get a load-hour 
curve as 00' . On the'other hand, if we plot (N) vertically 

N 




Fig. 43- 

as before and the service modulus (5) horizontally as in Fig. 
43, then the area under the curve thus obtained equals (M). 
This we can prove as follows. The area (A) under the curve is 

dA = NdS. 
dM 
AT 

or dM = NdS, 

so that dA = dM, 

and A = M, as stated. 



But 



dS 



REPLOT 



217 



The N-S diagram is of special interest because it shows at 
once the load, hours of service, and service modulus. 

Ordinarily the load curve is given as in Fig. 44A. This is 
the load curve as it is obtained from a graphical recording watt- 
meter, the horizontal distance being the calendar time. But 



7.000 



5000 

4000 

3000 

7000 

6000 

A 

5000 

4000 



2000 



1000 






M w^ 


■ 1*-^*^ B 


^ — -J 


1 3200) 




RUhr- SUM 


ItittWirp r -rf \\\\ 


ft I + Ii&f^^^^^^M.- 1-1 


1 III "II f iTOmff]^ 1 if 1 'J 


Mr A 




V 



730 1460 2190 2920 3650 4380 5110 58.40 6570 7300 8030 8760 
Hours 

Fig. 44. — Replot of a typical power plant load. 

to obtain even a roughly approximate equation between the 
load (M) and the time (N) would be impracticable for such a 
curve even if it were not impossible. 

For use in cost analysis, a replot is necessary in order to 
simplify the curve. The replot is made in descending values 
of (M). In order to do this, first plot on the ordinate (OM), 
Fig. 44B, the point (Mo) being the maximum instantaneous 
value of the load (M) for the year. Then draw a hori- 
zontal line as AB, in Fig. 44 A, at a height (Mi). Then the 
number of hours that the system operates at a load M± or 
over is given by the sum of the intercepts as shown, having 
a total value of (2Vi). With coordinates (Mi) and (2Vi) we 
can now plot the point (P) in Fig. 44B. By shifting the 
horizontal line AB, Fig. 44A, we can thus get as many points 



218 UNIT COST DETERMINATION 

as we desire on the replot of Fig. 44B and thus determine 
this curve. The equation for this is comparatively very 
simple, a common algebraic equation of the third degree 
usually giving a sufficiently close approximation. In Fig. 45, 



10 20 30 40 50 60 70 80 90 100 110 120 
Fig. 45. — Load curve and replot of the Portland Central Heating Co. 

we show the load curve and replot of the Portland Central 
Heating Co. from August to December 1914. 

137. Example 44. — For a certain power plant, the load 
hour equation is found to be 

M = 12,000 — o.oooi5A r2 . 
The cost per kw. of the installation is $70 and the fixed charges 
10%. If the operating costs are 0.2 cent per kw.h. deter- 
mine the total cost per kw.h. 

Solution: Here dM = — 0.0003^^, 





























> 




















A 


r 


/ 




VRe 


Blot 
















\ 


[\ 






















( 





) 




Q 
















S\ 


f 


-Load 


(Snrv< 





















\ 


/ 








w 

2 

u 

S3 
















\ 


/ 








PQ 


































Sept.- 


> 


e 


-Oct, 






-MtiVr- 



































































and 

so that 
When 



-/ 



dM 

N 



= — 0.0003 



J N 3 



S = - o.ooo3A r + B. 
N = 8760 hrs., 



(a) 



then 
for which 



EXACT UNIT COSTS 219 

M = 12,000 — 0.00015 (8760) 2 = 490 kw., 
5 = if. = o.o S 6. 

876O 



Substituting these values in equation (a) above, we get 
0.056 = — 0.0003 X 8760 + B 
B = 0.056 + 2.628 = 2.684. 

So that equation (a) becomes 

S = 2.684 ~ 0.0003N, 

and the cost per kw.h. is 

_ 2.684 — o.ooo^TV 

K 2 = - — 77 ^— 70 X 0.10 + 0.002, 

M 



or 



18.788 - 0.0021N , 

K 2 = - — + 0.002, 

12,000 — 0.00015^2 



We may tabulate values of the cost per kw.h. for variou: 
values of N, as follows: 



TABLE 91 



N s 


Ki 


.V 


K 2 





. 0035 


5000 


O . 00300 


1000 


0.0034 


6000 


0.00294 


2000 


. 0033 


7000 


O.00288 


3000 


0.0032 


8000 


0.00283 


4000 


0.0031 


8760 


O . 00280 



You will note that the cost of production of power is least 
during the periods of least load and greatest during the periods 
of greatest loads. If now the price or rate is made throughout 
proportionate to the cost, as it should and must be, the rate 
will be lowest during the periods of lightest load, thus encourag- 
ing the filling of the valleys in the power plant's load curve, 
reducing throughout thereby the cost and rate of the power 
service. 



22o UNIT COST DETERMINATION 



PROBLEMS 

1. A 10 h.p. pumping plant runs at full load for a variable number of 
hours per year. It costs $500 with 10% of fixed charges. The cost of opera- 
tion is 20 cents per hour. Determine the total annual and unit cost of 
production of the service. Plot the curves. 

2. A 25 kw. electric lighting plant carries a load of 25 kw. for 1000 hours 
per year and a load of 15 kw. for an additional 3000 hours per year. Deter- 
mine the total annual and unit cost of production of the power, if the plant 
costs $80 per kw. of capacity, with 12% fixed charges and operating costs 
of 1.2 cents per kw.h. delivered. 

3. A power plant cost $60 per h.p. installed, with fixed charges of 12%. 
The operating costs are 0.3 cent per h.p.h. The plant carries a load of 
1800 h.p. for 400 hours per year. During this 400 hour period the 
maximum demand is 2200 h.p. During the balance of the year, the plant 
carries a mean load of 600 h.p. with a maximum demand of 700 h.p. 
Determine the total annual, and the unit cost of production of the service 
during each period. 

4. A 100 mile branch railroad is built having a capacity of 200 cars of 
freight per day, and costing $2,000,000. The fixed charges are 13%. The 
cost of operation is 15 cents per car-mile loaded and 10 cents per car-mile 
empty. During two months in the year, the system carries an average of 
180 loaded cars out, with a maximum of 200, and an average return of 50 
loaded cars with a maximum of 60. During the. balance of the year, an 
average of 80 cars per day is carried out to the main line loaded, with a 
maximum of 100 and there is an average return of 20 cars with a maxi- 
mum of 30. Cars cost $3000, locomotives $20,000, with a capacity of 20 
cars. The weight of each locomotive is the same as a loaded freight car. 
Determine the total annual and unit cost of service per car-mile during 
each period. 

5. A steam engine generating plant costs $75 per kw. of capacity, with 
10% fixed charges. It carries a mean load of 5000 kw. for 60 hours per 
year, with a maximum of 8000 kw., a mean load of 3000 kw. for 1500 hours 
per year with a maximum of 4000 kw. and a mean load of 1000 kw. for the 
balance of the year, with a maximum of 1500 kw. The cost of operation 
is 0.1 cent per kw.h. with a mean load of 5000 kw., 0.12 cent per kw.h. 
when the mean load is 3000 kw.h. and 0.14 cent when the mean load is 
1000 kw. Determine the total annual, and unit cost of production of the 
service during each period. 

6. What could we afford to pay per kw. of capacity for a turbo-generating 
plant to take the place of the steam engine generating plant of problem (5), 
if its costs of operation are, for the mean load of 5000 kw., 0.12 cent, for 
the mean load of 3000 kw., 0.14 cent and for the mean load of 1000 kw., 
0.17 cent, other things being equal? 



PROBLEMS 221 

7. A power plant has an annual load-hour curve as given by the 
equation 

M = 5000 — 20 N + 0.0024 N 2 

and the cost of operation per h.p.h. is 0.2 cent. The first cost of the plant 
is $50 per h.p. of capacity installed, with 12.5% fixed charges. Determine 
the total annual, average unit, and true unit cost of production. Plot the 
curve of the variation of true unit cost with the variation of the load. 

8. A power plant carries a load as given by the equation 

M = 50,000 — $N. 

The first cost of the plant is $40 per h.p. of capacity, installed, with 10% 
fixed charges. The cost of operation per h.p.h. varies according to the 
equation 

A = 0.3 — 0.00004 M. 
Determine 

(a) The total annual cost of production per h.p. 

(b) The average cost of production per h.p.h. 

(c) The true cost of production per h.p.h. 

(d) Plot the curve of the variation of the true cost of production of the 
service per h.p.h. with the variation of the load. 



CHAPTER VII 

DETERMINATION OF SIZE OF SYSTEM FOR BEST 
FINANCIAL EFFICIENCY 

Variable Operating Cost. Total Production Cost. Determination of 
Equation of Actual Load Curve. Analysis of Heat Transmission. 
Point of Best Financial Efficiency. Determination of Size of System 
for Best Financial Efficiency. 

138. In the past we have assumed that the operating cost 
per h.p.h. was a constant. This is far from true, especially 
in smaller sized plants and this consideration must modify 
our conclusions considerably where best economy is aimed 
at. So also within any large plant, the unit operating cost 
at full load will be considerably less than at fractional loads. 

If then we call a h ch, a s , etc., the operating costs, during 
given periods and corresponding loads, we can determine our 
costs as before, but modified by this variation in operating 
cost. 

For the first period, then, the total annual cost Kn is 
K n = Mtcp + L&iNi, 
and the cost per h.p.h. is * 

_M 1 cp 

During the second period, the total cost for the excess load is 
Kn! = M 2 cp + L 2 02 f N 2 , 

where a 2 is the excess operating cost per h.p.h. required over 
that of the first period. 

The cost of operating the unit L\ for the N 2 hours is 

K 12 " =^cpN 2 + L iai N 2} 



VARIABLE OPERATING COST 223 

so that the total cost of the second period is the sum of K x2 
and Kn", or 



K n = (M± N 2 + M 2 \ cp + (Ltd* + L 2 a 2 )N 2 . 



(95) 



C P 

+ a. 



If we now call the mean operating cost for the load 
(Li + L 2 ) then 

(Liai + ^202) = (Li + L 2 )a, 

and Kn = (Ml N 2 + M 2 \p + (I* + L 2 ) aN 2 . 

The cost per h.p.h. during this period is then 
_(M 1 MA_ 

So in general 

K 2 n = O n * yy- + 0n, 

where L n r = L x + L 2 + • • • + L n , 
and the service modulus 

6n "iVi + iV 2 + * ' ' + AV 

And so, also, when, the length of the period becomes in- 
fmitesmal, then 

where 5* = / — as before, and (0) is variable. 

Knowing then M = /(A), and = F{M) either by exact or 
approximate equation, a complete solution of the costs can be 
made, either of an existing system, or in advance of its 
construction. 

139. One of the advantages in using instantaneous costs 
per h.p.h. as given by the equation 

K 2 =S C 4+a 
M 



224 SIZE OF SYSTEM FOR BEST FINANCIAL EFFICIENCY 

is that the conditions for minimum costs can be determined 
by the usual method of equating the first differential to zero, 
and the system designed to meet these conditions. 

If in the cost determination, the entire system is treated as 
a unit, so that the costs include the transmission and distribu- 
tion as well as plant costs, then the above determination for 
minimum costs will determine both the size of plants and the 
corresponding areas of distribution for any given set of con- 
ditions, such that maximum financial efficiency will be attained, 
i.e., so that the service will be rendered at least cost. 

In determining the best size of a system for minimum cost, 
we must consider the unit first cost (c) of the system as vari- 
able, since this varies with its size. Once this size is deter- 
mined, (c) becomes fixed and has a definite constant value. 

Since the instantaneous production cost per unit of size 
(h.p.) per hour is 

the cost per hour for the size (M) will be 

K 2 ' = Sep + Ma, 
and the total production cost per year will be 

/*876o /•8760 

K = p ScpdN+ / MadN . . . (96) 

Jo Jo 

The condition for minimum total production cost is then 

dK = 

dM " °* 

The operating cost varies not only with the size of the units 
but also with the per cent of full load that they are carrying, 
because their efficiencies vary with fractional loads often very 
greatly. The handling of the entire subject is quite complex 
but this should give peculiar satisfaction to engineers for it 
insures that in future the matter must be taken out of the 
hands of amateurs. 

Example 45. — A power plant carries a load as shown in 
Fig. 46. If the cost of the plant per h.p. is $50, fixed charges 



MINIMUM PRODUCTION COST 



225 



are 12%, and the operating cost per h.p.h. is 0.8 cent, 
determine . 

(a) The cost per h.p.h. 

(b) The total annual cost. 

Solution: The first step necessary is to determine the equa- 
tion of the curve shown in Fig. 1. In order to do this, we as- 
sume the equation to be 

M = M + AN + BN 2 + CN* + DN* . . . . (1) 

The longer the equation, i.e., the more constants, the 
more nearly we can approximate to the true curve. For each 













































6000 


















































































5000) 


















































































4000 


















































































3000 


















































































2000 


















































































1000 












































































X 

















































1000 2000 3000 4000 5000 6000 7000 8000 8760 

Hours 

Fig. 46. — Load curve for example 45. 

constant in the equation we can make the approximate curve 
run through one point on the true curve. Thus in the above 
equation a there are five constants, and we can therefore make 
the approximate curve run through five points on the true 
curve given. If we had another term (EN 5 ) then we could 
make it run through a sixth point and therefore approximate 
to the true curve just so much more closely. 

If we express (M) in thousands of kilowatts and (N) in thou- 
sands of hours, then we can choose the following five points 



226 SIZE OF SYSTEM FOR BEST FINANCIAL EFFICIENCY 

of the true curve through which to make the approximate 

curve as given by equation i run. Thus, when 

N = o, M = 5, 

N = 2, M = 4 , 

N = 4, M = 2, 

N = 6, M = i, 

N = 8, M = 0.9. 

Substituting in equation 1, the value N — o and M = 5, 
we get M = 5, so the equation becomes 

M = 5 + AN + BN 2 + CN* + DN\ . 4 (2) 

Our task is now to determine the values of A , B, C, and D. 
We shall follow this through in considerable detail. 

Substitute in equation 2 the value N = 2 and M = 4, and 
we get 4 = 5 + 2^+ 4 #+ 8C + 16D, 

or o = 1 + 2 A + 4B + 8C + 16D .... (3) 

Then substitute in equation 2 the values N = 4 and M = 2, 
and we get 

2 = 5 + 4,4 + i65 + 64C + 256^ 
or = 3+4-4 + i6£ + 64C + 256D ... (4) 

The substitution in equation 2 of the values N = 6, M = 1 
gives o = 4 + 6^4 + 36.B + 216C + 1296Z}, 

or o = 2 + $A + i85 + 108C + 648Z). . . (5) 

And finally the substitution of the values N = 8, M = 0.9 
.gives o = 4.1 + 8^4 + 64.8 + 512C + 4096D . (6) 

We have thus the four following equations: 

o = 1 + 2 A +4B +SC + 16D (3) 

o = 3 + 4A + i6£ + 64C + 256D ... (4) 

= 4 + 6^+ 36^ + 216C + 1296Z) . . (5) 

and o = 4-1 + 8^ + 64B + 512C + 4096D . (6) 

We can now eliminate {A) between equations 3 and 4 by mul- 
tiplying equation 3 by 2 and subtracting from equation 4, 
thus o = 2 + 4A + SB + 16C + 32Z) 

o = 3 + 4A + 16B + 64C + 256D . . . (4) 
subtracting, = 1 + SB + 48C + 224Z) ... (7) 



LOAD-HOUR CURVE 227 

We can also eliminate (A) between equations 3 and 5 by 
multiplying the former by 3 and subtracting from the latter, 
thus 

o = 3 + 6A + 12B + 24C + 4 80 
o = 4 + 6,4 + 36^ + 216C + 12960 
subtract = 1 + 24$ + 192C + 12480 . . (8) 

And finally, we can get a third equation with (^4) eliminated, 
by multiplying equation 4 by 2 and subtracting from equation 
6, thus = 6 + 8,4 + 325 + 128C + 512Z) 

o = 4-1 + 8,4 + 64.B + 512C + 40960 
subtract o = — 1.9 + 32.B + 384C + 35840 (9) 

This gives us the three following equations: 

o = 1 + SB + 48C + 224D (7) 

= 1 + 24B + 192C + 12480 (8) 

o = - 1.9 + 32JB + 384C + 35840. • . (9) 
We now proceed to eliminate (B) from these equations. 
Multiply equation 7 by 3 and subtract from equation 8. This 
gives = 3 + 24B + 144C + 672Z) 
= 1 + 24B + 1Q2C + 12480 
subtract o = — 2 + 48C + 5760 (10) 

Also multiply equation 7 by 4 and subtract from equation 9. 
This gives o = 4 + 32.8 + 192C + Sg6D 

o = - 1.9 + 32I? + 384C + 35840 
subtract o = — 5.9 + 192C + 26880 . . (11) 

We have now only the two equations following, with only 
two unknowns, C and 0. 

o = - 2 + 48C + 5760 (10) 

o = - 5.9 + 192C + 26880. 

Multiply equation 10 by 4 and subtract from equation 11. 

This gives o = — 8 + 192C + 23040 

o = - 5.9 + 192C + 2688Z) 

subtract o = — 2.1 + 3840 

2 1 
So that = — +— = — 0.00547. 

3 8 4 



228 SIZE OF SYSTEM FOR BEST FINANCIAL EFFICIENCY 

By substituting this value of D ia equation 10, we get 

O = - 2 + 4 8C - ~ X 576, 

3 8 4 
or o = - 2 + 48C - 3.15. 

So that C = ^~ = 0.1073. 

48 

We have now the values of both C and D. Substituting 
these in equation 7 gives 

o - x + SB + ^X5^5 _ 224 x « 

48 384 

or o = 1 + SB + 5.15 — 1.225. 

So that o = SB + 4.925, 

and B = - ±^$ = - 0.6156. 

8 

Dividing equation 3 by 2 gives 

o = 0.5 + A + 2B + 4C + 8Z), 

and substituting the values found for B, C, and D gives 

o = 0.5 + A - 1. 23125 + .429167 - .04375, 
or o = 0.929 — 1.275^4, 

so that A = 0.346. 

Substituting these values in equation 2, we get 
M = 5 + 0.346N — 0.61 562V 2 + 0.1073N 3 — 0.00547N 4 (12) 

as the approximate equation of the curve. The crosses (x) in 
Fig. 46 indicate points calculated from this equation. The 
conformation is very close. 

We can now determine the service modulus (S), since by 
equation 12 

dM = (0.346 - 1.2312N + 0.3219N 2 - o.o2i8&N*)dN, 

„ _ n^°dM 

6 " J* ~N 7 

p- 76 ° (0.346 - 1.2312^ + 0.3219N 2 - o.Q2i88A^ 3 ) dN , 
6 ~J N N 



LOAD-HOUR CURVE 229 

or 5 = 0.346 \og e N — 1.2312.Y+0.1610T 2 — o.oo729A 73 + C. 1 (13) 
Where C is a constant of integration. By equation 12, when 
N = 8.760, M = 0.71, so that when N = 8.760 

5 = §1± = 0.08105. 
8.760 

Substituting this value of (5) and of (.V) in equation 13, 
gives 

0.08105 = 0.346 X 2.1702 — 1. 2312 X 8.76 + 0.161 

X (8.76) 2 - 0.00729 X (8.76) 3 + C 1 , 
so that C = 2.6612 

and 
5 = 0.346 \og e N — i. 23i2A 7 +o. i6iiV 2 — o.oo729iV 3 + 2. 6612 (14) 

The cost per kw.h., is 

M 

S S 

or K = 50 X 0.12— - + 0.008= 6 — + 0.008. 

M M 

Substituting the values of (5) and (M) found, we get 

K _ 6(0.346 log e A" — i.23i2A r + 0.161A 72 — 0.00729^ + 2.6612) 
1000(5 + 0.346A 7 — o.6i56A r2 + 0.1073A 73 — 0.00547A 74 
+ 0.008 (15) 

as the cost per kw.h. in terms of N. (Ans. a.) 
The total cost per hour for the load (M) is 



K 1 = 6ooo5 + 0.008M X iooo 2 = 6ooo5 + 8ooolf, 

where (M) is expressed in thousands of kilowatts and N in 
thousands of hours. 

The total production cost per year is 

/"8.760 /*8. 7 6o 

Ki= MK'dN= 6000 / SdN 

Jo Jo 

/8.760 
MdN. . (16) 



230 SIZE OF SYSTEM FOR BEST FINANCIAL EFFICIENCY 

The integral 

fsM.f/f <»./<§»» -/->.<*, 

,iur\ 5222' 

= (M)o = 500c 

since dS = -^r 

N 

The total fixed charges per year are 

6 I S dN = 6000 X 5000 = $30,000. 

The number of millions of kilowatt-hours (thousands of kilo- 
watts multiplied by thousands of hours) produced per year is 
given by 

(5 + O.346N - 0.6I562V 2 + 0.1073^ - 0.00547^)^ 

= (5A^+o.i73iV r2 -o.2052xV 3 +o.o2682A' 4 -o.ooio94^ 5 ) 8 - 76 
= 5 X 8.76 - 0.173 X 76.7376 - 0.2052 X 672.2214 
+ 0.02682 X 5888.6595 — 0.001094 X 51584.66 
= 43.80 - 13,276 - 137-94 + 157-934 - 56.434 
= 20.636 millions of kilowatt-hours, costing 
= 20.636 X 8000 =.$165,088 

whence the total annual cost is 

30,000 + 165,088 = $195,088. (Ans. b.) 

Example 46. — If in the above example the operating cost 
per thousand kw.h. is given by the equation 

a = (10 — M) 

between 500 and 5000 kw., where M is again expressed in 
thousands of kw., 

(a) Determine the cost per kw.h. and 

(b) Determine the total annual cost. 

Solution: In the previous problem, the operating cost (A) 
was assumed constant. In this problem, it is assumed variable 



TOTAL PRODUCTION 231 

in accordance with the above equation. It is evident that the 
value of (S) , the service modulus, will not be affected by varia- 
tions in the operating cost. Equation 15 above will not be 
changed except in the last term, which must be changed to 
the new value. We therefore get, for the cost per kw.h. ( K) 
K _ 6(0.346 log e 2V — 1.2312N -fo.i6i2V 2 — 0.007292V 3 + 2.6612) 
1000(5 + 0.3462V — 0.6156A 2 + 0.10732V 3 — 0.005472V 4 ) 

. (10-M) /A x 

+ i J - (Ans. a), 

1000 

where as before 

M = 5 + 0.3462V - 0.6156A 2 + 0.1073A 3 - 0.00547A 4 . 

The total cost per year is then 

/•8.760 /•8.760 

K x = 6000 / SdN + iooo 2 / aMdN. 

%) o Jo 

/8.760 
S dN = $30,000 as before 

J/^S.760 /"8.760 

' aMdN = 1000 / (10 - M)MdN 
o Jo 

/8.760 
(10M - M 2 )dN. 

Substituting for M its value in terms of (2V) as given above, 
we get 
1000/(25 - 0.12A 2 + 0.426N 3 - 0.45272V 4 + 0.13592V 5 

— 0.01825 A 7 " 6 + 0.0011742V 7 — 0.000032V 8 ) d2V 
= 1000(25^ — 0.042V 3 + 0.10652V 4 — 0.090542V 5 

— 0.022652V 6 — 0.00261A 7 + 0.0001472V 8 

— o.ooooo33A 9 ) 876 ° 

1 
= iooo2V(25 — 0.04A 2 + 0.10652V 3 — 0.090542V 4 + 0.03365M 

— 0.002612V 6 + 0.0001472V 7 — o.ooooo332V 8 ) 8 - 76 
= 8760(25 - 3.07 + 71.6 - 533.0 + 1166.1 - 1177.1 

+ 580.7 - 115.4) 
= 8760 X 14.2 = $124,392 operating costs, 
so that the total annual production costs are 

Ki = 30,000 + 124,392 = $154,392. (Ans. b.) 



232 SIZE OF SYSTEM FOR BEST FINANCIAL EFFICIENCY 

While, as may be seen from the above, the work of solving 
such cost problems requires skill and considerable painstaking 
effort, it certainly permits of real and conclusive deductions. 

140. In the following problem of determining the size of a 
heating system for best financial efficiency, we have assumed 
the heat to be transmitted in water instead of steam. It may 
be of interest to note the comparative cost of transmitting 
heat energy by steam and by water. 

We can allow approximately a velocity of ioo ft. per second 
for steam and 5 feet per second for water, the velocity of steam 
being twenty times as great as that of water. The heat capacity 
of water is one B.t.u. per pound, or 62.5 B.t.u. per cu. ft. The 
heat capacity of steam varies with the pressure (density) in 
accordance with the following table. 



TABLE 92 

Heat Capacity for Steam above 150 F. 



Pressure, 


Temp., 


Heat 


Heat per cu. ft. 


Heat per cu. ft. 


lbs. abs. 


dry sat. 


per cu. ft. 


per degree F. 


per degree X 20 


10 


193 


26.5 


0.62 B.t.u. 


12.4 


15 


213 


39° 


O.62 


' 


12.4 


20 


228 


51-5 


0.66 


' 


13.2 


50 


281 


124 


0.95 


' 


19.O 


IOO 


328 


240 


i-35 


' 


27.0 


150 


358 


357 


1. 71 


' 


34-2 


250 


401 


583 


2.32 


< 


46.4 


350 


432 


814 


2.90 " 


58.0 



The third column is obtained by dividing the total heat per 
pound of the steam above 150 F. by the number of cubic feet 
per pound of steam, thus getting the B.t.u. per cu. ft. The 
fourth column is then obtained by dividing the heat per cu. 
ft. by the temperature of the steam above 150 . But inas- 
much as steam may be allowed to travel 20 times as fast as 
water, we have multiplied this column by 20, giving the last 
column. This shows 12.4 to 58 B.t.u. for steam as compared 
with water at 62.5 B.t.u. per cu. ft. per degree. A number of 
important factors must be borne in mind. If the pressure 



TYPE OF HEATING SYSTEM 233 

is over ioo#, then extra strong pipe and fittings must be used. 
Since these are far more expensive than standard, the cost of 
transmission is greatly increased, which does not permit of a 
fair comparison with that of hot water at nominal pressures. 
Therefore the real capacity of steam is really only 12.4 to 27 
B.t.u. as compared with water at 62.5. 

On the other hand, while water transmission requires as 
large a return pipe as outgoing pipe, that for steam may be 
very small, and if the condensate is wasted, the return pipe 
may be done away with altogether. But in the latter case, 
the cost of purchasing or pumping the water thus wasted must 
be added to the total service cost. 

But even without a return pipe for the steam, and allowing 
nothing for the cost of water thus necessarily wasted, we have 
12.4 to 27 B.t.u. for steam as compared with half of 62.5, or 
31.25 B.t.u., for water transmission. So that even under these 
conditions the cost of steam transmission is at least 16% 
greater than for water. The cost of water wasted usually 
runs from one to ten cents per 1000 lbs., a fairly significant 
item itself. 

The conclusion cannot be avoided, then, that in spite of the 
fact that most central heating stations are designed for steam 
transmission, heat transmission by water is far more economical. 
If, in the hot water system, instead of heating water in the 
boilers, we generate steam, using the steam for the generation 
of power, and using the exhaust steam (atmospheric pressure, 
212 ) for the heating of the water, we have an incomparably 
more economical system. 

Example 47. — Assume that the cost per bo.h.p. of a central 
heating plant varies as curve A, Fig. 47, and that the cost of 
transmission pipe per foot complete in place varies as curve 
A, Fig. 48, and that the operating cost is one cent per 
bo.h.p. hour, neglecting radiation loss in transmission, deter- 
mine the area of transmission for maximum economy; in blocks 
of 200 ft., if the maximum demand is 268,000 B.t.u. per block 
per hour and the mean annual load 67,000 B.t.u. per block 
per hour, fixed charges 10%. 



234 SIZE OF SYSTEM FOR BEST FINANCIAL EFFICIENCY 



$120 



100 



80 

EC 
1 60 



40 



20 



.100 200 300 400 500 600 700 800 900 1000 

Bo. H.P. 

Fig. 47. — First cost per horse power of heating plant assumed in example 47. 



A 













































$2.40 


















































































2.00 


















































































a 

gl.60 

M 


















































































|l.20 

4-> 
01 


















































































5 

0.80 


















































































0.40 





























































































































36 



4 8 12 16 20 24 28 32 

Cross-sectional Area of Pipe 

Fig. 48. — Variation of cost of pipe with size assumed in example 47. 



40 



SIZE OF HEATING SYSTEM 235 

Solution: The total production cost per year is 

Ki = Mcp + LaN, 
where p = 0.10, 
a = 0.01, 
N = 8760. 

Assuming the shape of the area served to be square, the 
number of blocks on each side being (B) blocks from the plant, 
then the total number of blocks served will be 2B(B + 1), 
in which B is always odd. Since each block requires 268,000 
B.t.u. maximum, or 8 bo.h.p., the total maximum load will be 

M = 2B (B + 1) X 8 = 16B (B + 1) bo.h.p., 
and the mean annual load (L) will be 

L = i X 16B (B + 1) = 4B (B + 1) bo.h.p. 
whence the annual operating cost will be 
LaN = 4B (B + 1) X 0.01 X 8760 = 350.4.5 (B + 1) dollars. 

We must now determine (C), the cost of the system per 
bo.h.p. Curve A, Fig. 47, gives (G) the cost per bo.h.p. 
of the heating plant. So we must only determine (C 2 ), the 
cost per bo.h.p. of the transmission according to the cost 
data given in Fig. 48. 

Allowing 200 F. as the maximum temperature of the out- 
going water and 150 as the temperature of the return water 
we can carry 50 B.t.u. per lb or 

50 X 8| = 417 B.t.u. per gallon. 

Since each block requires 268,000 B.t.u. maximum per hour, 
or 4470 B.t.u. per minute, there will be required 

4470 -T- 417 = 11 gallons per minute per block. 

This will require, for a friction head of about one foot per 
hundred feet of pipe, a ij" pipe. 

The length and size of distribution pipe may be computed 
as follows: 

(1) For one square block transmission area 6 

Pipe required 4 X 1 (/) of 11 g.p.m. capacity, where / = 200 ft. 

(2) For a square of 3 blocks on a side 

Pipe required is 4 (/) (1 + 3) of 11 g.p.m. capacity. 
4 (/) X 2 of 22 g.p.m. capacity. 



236 SIZE OF SYSTEM FOR BEST FINANCIAL EFFICIENCY 

(3) For a square of 5 blocks on a side 

Pipe required is 4 (Z) (1 + 3 + 5) of 11 g.p.m. cap. 
4 (I) X 4 of 22 g.p.m. cap., 
4 (0 X 2 of 44 g.p.m. cap. 

(4) For a square of 7 blocks on a side 

Pipe required is 4 (I) (1 + 3 + 5 + 7) of 11 g.p.m. cap. 
4 (/) X 6 of 22 g.p.m, cap., 
4 (/) X 4 of 44 g-P-ni. cap., 
4 (I) X 2 of 66 g.p.m. cap. 

(5) In general for a square B blocks on a side 

Pipe required is 4 (I) (1 + 3 + 5 + 7 + 9 + • * • + B) ft. of 
11 g.p.m. cap., 

4 (I) X (B — 1) ft. of 22 g.p.m. cap., 
4 (0 X (B - 3) ft. of 44 g.p.m. cap., 
4 (/) X {B — 5) ft. of 66 g.p.m. cap., 
4 (0 X (5 - 7) ft. of 88 g.p.m. cap., 
etc. 
The cost of the i|" pipe according to Fig. 48 is 34 cents per 
foot, and since the cost in place is assumed in direct propor- 
tion to the area of the pipe, the cost per foot of the 22 g.p.m. 

pipe is 

34 X 2 = 68 cents per ft.; 

of the 44 g.p.m. pipe is 

34 X 4 = $1-36 per ft., etc. 

The total length of ij' r pipe required is 

4 X 200 (1 + 3 + 5 + 7 + • • • + B) ft., 

800 X (^T^Y = 2 °° ( B + J ) 2 ft -> 

and its cost is 

200 (B + i) 2 X 0.34 = 68 (B + i) 2 dollars. 
The cost of the 22 g.p.m. pipe is 

4 X 200 (B - 1) X 0.68 = 544 (B - 1) dollars; 

of the 44 g.p.m. pipe, it is 

4 X 200 (B - 3) 2 X 0.68 = 544 X 2 (B - 3). 



SIZE OF HEATING SYSTEM 237 

Of the 66 g.p.m. pipe, it is 

4 X 200 (B - 5) 3 X .68 = 544 X 3 (B - 5), 
etc., up to 

4 X 200 X 2 X ( — — J X .68 = 544 ( — — J X 2, 

$7000 

11 Gpm. 



__,._ ._ - --J- .--- -; 




























































\ 7 


: ■•: : s,. ji. • 


S,. -,Z 






::_:._ 2i^s t: 


i J^%z - 


-,Z s • 


.,z s^ 


. . /. s ., 































































Fig. 49. — Heat distribution layout for example 47. 

whence the total transmission cost is 

C 2 = 68 (B + 1)2+ 544 {B - 1) + 2 (B -3) + 3 (5 - 5) 



+ 4 (5 - 7) + 5 (B - 9) + 



+ 



Letting 



m 



X 2. 



J-[(B-i) + 2 (B- 3 )+3(B-s)+- • ■+(^ Ji ) X2 ] 

we set 

C 2 = 68 (B + i) 2 + 544/. 



238 SIZE OF SYSTEM BEST FOR FINANCIAL EFFICIENCY 

Dividing through by 

M = i6B(B + i) 
gives the cost of transmission per bo.h.p., or 

r>- ,': A B+l) i 34/ 

C2 — 4.2s h — : " • 

5 B B(B + 1) 

The variation of (J) and C 2 ' with (5) is as follows: 
B 1 3 5 7 9 11 13 15 17 19 21 

/ O 2 8 20 40 70 112 168 240 33O 440 

C 2 ' 8.5 11.3 I.42 17.O 19.8 22.6 25.6 28.3 31.2 34.O 36.8 

Assuming now that we can represent the cost of the heating 
plant (Ci) per bo.h.p. by the following equation between 600 
and 1200 bo.h.p., 

G = 61.66 — 0.035M + o.cccoii33i4' 2 . 
We get, since 

M = i6B(B + 1), 
that d = 61.66 - 0.035 X i6B(B + 1) 

+ 0.00001133 X 16 X i6B 2 (B + i) 2 
or ' Ci = 61.66 - 0.56^(5 + 1) + o.oo2 9 B\B + i) 2 . 
Where the total first cost (C) of the system per bo.h.p. is 

C = d + C 2 ' } . 
or C = 61.66 - o.s6B(B + 1) + o.oo2 9 B 2 (B + i) 2 

+ 4.2 5 ( ^ + l) + 34/ , 
^ 4 5 B ^ B(B + i) 

the annual fixed charges per bo.h.p. are 

Cp = 6.166 - 0.056^(5 + 1) + 0.0002 9 5 2 (J5 + i) 2 

+ o, 425 (*±i) + 3V , 

Since the total operating costs per year are 

LaN = 35oaB(B + 1), 
the cost per installed bo.h.p. 

LaN 35oaB(B + 1) = 
M i<SB(B + 1) 9 ' 



SIZE OF HEATING SYSTEM 



239 



the total production cost per bo.h.p. per year is therefor 
K = 21.9 + 6.166 - o.o$6B{B + 1) 

+ o.ooo2 9 B*(B + i) 2 + 0.425 ( ~- L) + B{ 3 B 4 l iy 

or K = 28.066 - o.os6B(B + 1) + o.ooo2q£ 2 (5 + i) 2 

3AJ 



{B + i) , 

+ O.425 -— ^~ 



B ' B(B + 1) 
Differentiating (K) with respect to (B) and equating to 
zero, will give very complex results. We can, however, make 
a solution by plotting. Thus 



B 


1 


3 


5 


7 


9 


11 


13 


15 


17 


19 


21 


K 


28.7 


29.1 


28.1 


27-S 


27.4 


28 


30 


32.8 


41.2 


52 


67.8 



The maximum financial efficiency evidently takes place at 
B = 9, an area of service of 180 blocks. Under this condition 
M = i6B(B + 1) = 16 X 9 X 10, 
M = 1440 bo.h.p. 

The curve for the variation of the production of service cost 
per bo.h.p. per year with the length of area served is given in 



21 

a 19 
< 17 
J 15 

1 13 
I 11 



x 



$30 



$50 



$40 
Cost per Bo.H.P. Year 

Fig. 50. — Variation of the unit cost of service with the size of the system 

according to example 47. 



$60 



2 4 o SIZE OF SYSTEM BEST FOR FINANCIAL EFFICIENCY 

Fig. 50. Note that after the point of best economy is passed 
the increase in cost is very rapid. The cost of service of a 
heating system that was considerably too large would be far 
in excess of the individual plant costs. However, the results 
obtained above must not be generalized, as they apply only 
under the conditions assumed in the problem. But under 
these conditions, a steam heating system is still more expen- 
sive and capable of serving only a smaller area economically. 
Another factor that must be borne in mind is load density. 
We have in the above problem assumed uniform load density. 
The load density may be very much greater or less than that 
assumed, while on the other hand the load density, instead of 
being uniform, may vary over the area in a great many essen- 
tially different ways. The effect of these variations on the 
system's size and economy afford study of the most valuable 
kind in practice. 

PROBLEMS 



1. Assume the annual load curve to be as in Fig. 51, the first cost of 
the power plant to be $60 per h.p. installed, fixed charges 12%, and 
operating costs to be 0.6 cent per h.p.h. 



5000 



4000 



^3000 



2000 



1000 













































































































































































































































































































\ 


s s 








































































































































































> 


\ 







JL000 2000 3000 ioOO 5000 6000 7000 8000 87&0 
Hours 

Fig. 51. — Load curve for problem 1. 



PROBLEMS 241 

(a) Determine the total annual cost of production, 

(b) Determine the cost of production per h.p.h. and 

(c) Plot the curve of variation of cost per h.p.h. with the load. 

2. If in problem (1), the operating cost in dollars per h.p.h. is given by 
the equation 

a = 0.01125 — 0.00000125 M 

and all other things are equal, determine 

(a) the total annual production cost, 

(b) the number of h.p.h. produced per year, 

(c) the cost of service per h.p.h. and 

(d) plot the curve of variation of cost of service per h.p.h. with the 
size of the load. 

3. In a power plant the load varies with time according to the equation 

M = 5OOO + 2N — 0.0002iV 2 , 

and the cost of operation per kw.h. varies according to the equation 

a = 0.015 — 0.000001M. 
The first cost of the plant per kw.h. is $75, fixed charges 10%. Determine: 

(a) the total annual production cost, 

(b) ' kw.h. produced per year, 

(c) " average cost per kw r .h., 

(d) " true cost per kw.h., 

(e) " maximum cost per kw.h. and 

(f) " minimum " " kw.h. 



CHAPTER VIII 
DETERMINATION OF TYPE AND SIZE OF UNITS 

Stand-by Units. Number of Units in a Plant for Given Load for Best 
Financial Efficiency. Economy of Units at Fractional Loads. 

141. In considering the matter of the proper selection of 
the size of units in any given plant, too much emphasis has been 
placed on the necessity of a reserve unit. Like most things, a 
reserve unit is desirable in inverse proportion to its cost. 
When this cost rises above the probable gain, it is no longer 
good judgment to provide it. 

The first question is, how long will a given piece of ma- 
chinery run, without a shutdown, under conditions of aver- 
age care? While this is a very difficult matter to decide on, 
yet as a rule the number of hours of run without shutdown will 
be in direct proportion to the life of the equipment. But 
with this, we must consider the nature of the equipment as 
well. For example, the life of a motor is twice that of an 
engine, therefore we may expect twice as long a run. But as 
a matter of fact, we will get more than that, because the 
shock of reciprocating parts is absent in the motor. So as a 
whole, the period of continuous run is far greater for all rota- 
tive classes of machinery, such as generators, turbines, centrif- 
ugal pumps and the like, than for the reciprocating classes, 
as engines, compressors and such. 

That shutdown will occur is certain, for no machine is 
built to run continuously during its entire life, without certain 
parts being either adjusted or repaired, or entirely replaced. 
Among such parts are bearings, crosshead guides, stuffing boxes 
and the like. However, the period of necessary shutdown 
may be anticipated, and for that matter usually must be an- 
ticipated to avoid severe damage to the machine, due to break- 

242 



STAND-BY UNITS 243 

down. And its repair may therefore be accommodated to the 
load being carried. 

But this is not the primary consideration in the matter of 
dividing our load among a number of units. The menace of 
shutdown is not nearly so great as one would be lead to believe 
by the emphasis placed on it by so many authors. First-class 
machinery, run with average care, under normal loads, may 
be depended on to run at least a certain period without any 
danger of shutdown. They will then require certain repair, 
after which they may again be relied upon with reasonable 
certainty. 

142. The reason for the design and installation of any system 
is the rendering of a given service. The primary object for the 
subdivision of the plant into several units must be to render 
this service at as low a cost as possible. We must therefore 
take into consideration a number of primary factors that con- 
trol this. These are as follows: 

1. For larger units: 

(a) Decreased unit first cost of the equipment with in- 

creased size. 

(b) Increased efficiency under similar conditions for the 

large units. 

2. Against larger units: 

(a) Decrease in efficiency at fractional loads. The larger 
the unit, the smaller per cent of full load the unit 
will run at, under average conditions. 

For example, if the maximum load on a certain power plant 
is 10,000 kw. and the mean load is 3000 kw., then if we have 
only one unit, it will run on the average at only 30% load, 
with corresponding low efficiency. If on the other hand we 
put in two or more units, we can shut down one unit after an- 
other as the load decreases, maintaining much nearer full load 
conditions on those units that are kept in operation, and get- 
ting correspondingly much higher efficiency under actual 
running conditions. 



244 TYPE AND SIZE OF UNITS 

Calling {A) the attendance cost per h.p.h., 

(F) the fuel cost per h.p.h. at full load 
and (P) the per cent of full load efficiency at (X) per 

cent of full load 
then at full load, the operating cost (a) is 

a = A + F per h.p.h (98) 

and at fractional loads, the cost of operation is 

p 
a = A + j, (99) 

the load being under these conditions ( XM) . 

If we are actually carrying some load as may be given by 
some equation as 

N =f(M), 

then the operating cost per hour is 

(a+I)m, 

and the total operating cost per year is 

£ 76 \a+1)m<in 

and thus the production cost per h.p.h. is 
and the total production cost per year is 

/8760 /*8 7 6o / /A 

ScdN+ / I A +^-\MdN. (101) 

143. The above are the conditions for a single unit or where 
all the units are operated, each carrying a uniform proportion 
of the entire load. But it certainly is not good practice, where 
there are a number of units, to operate them all at small loads, 
when it would be possible to shut one or more of them down 
and operate the balance at more nearly full load. 

Assuming then that we have a number of units, Mi, Mi, 
Mz, ... in our plant all of the same size, and that we operate 
them so that all which are in operation will be at full load 



* 2 = if 4 



SERVICE COST 



245 



except one, then for any load (M) we will have, let us say, ( U) 
units of (Mi) size operating at full load and one at fractional 
load. Under these conditions the total attendance cost per 
hour is (AM) and the total fuel per hour is 

F 



UMiF + M } 



P 9 



where UM l + XM 1 = Mi( U + X) = M, . . (102) 

and (X) is the per cent of full load on the unit that is run- 
ning light and (P) is a function of (X), which is obtained as 
an approximate equation from the actual test (or factory 
guarantee) curves. Whence the total operating cost per hour 
for the load (M) is 

AM + UMxF + M 1 (j\ = AM + M,F (u+j\ (103) 



But since 



M x 



M 



U + X 

and P=f(X), 

the total operating cost is 



AU + w ! h-A u+ ) 



( v +m)- 



M 



A + 



( u+ ido) 



( u + X) J 



(104) 



and the operating cost per h.p.h. (a) is 



A + F- 



■ (105) 



(U+ X) 
So under these conditions, the unit production cost is 



Kt = S -$ + A + 

M 



( u + m) 



(U+ X) 

and the total annual production cost is 

F 

/8760 r a * 

ScdN+ I 



.(106) 



•8760 



M 



A 



(U+X) _ 



dN (107) 



M. 



246 TYPE AND SIZE OF UNITS 

If the size is already determined, then (c) is a constant and 
the first part of (K 2 ) above is simply 

pc.fsdN. 

But fsdN = f*NdS 

since both give the area under a curve the equation of which is 

5 = MN). 

So also dS = — — • 

N 

Whence f S dN = Cn dS= f % N d ^-= f dM 

Bearing this in mind will always simplify this part of the 
calculation. 

144. In the above, the fixed charges must be based on the 
total installation, while the operating cost depends on the units 
in operation. In practice we should have to determine the total 
annual operating costs (a) by a series of integration instead 
of only one. If, for example, we have ( U + 1) units in the 
plant, we would integrate from ( U + 1) to (U), then from 
U to ( U — 1), and so forth, i.e., always from X = 100% to 
X = o, or if the units will carry say 25% overload safely, 
and run with better economy than below 25% load, then our 
integration steps would have to be between X = 125% and 
X = 25%. The reason for this is that at the point of shut- 
down of a unit a point of discontinuity is introduced in our 
curve. 

It is evident then that units like generators, steam engines, 
and the like, that are technically capable of carrying overloads, 
should be built mechanically well to carry this overload for a 
reasonable per cent of their running time without giving 
trouble, in order to allow the operator to get the lowest pro- 
duction costs. Invariably better economy is obtained at 
reasonable per cent overload than near zero load. For ex- 
ample, a steam engine will use about \\% more steam be- 
tween full load and 25% overload, at \ load it will use 60% 



DETERMINATION OF NUMBER OF UNITS 247 

more steam, and below that still more. A steam turbine will 
use 1% more steam at 25% overload than at full load, while 
at i load it will use 15 to 40% more steam, and so on. 

In the determination of the number of units (of equal size) 
required in a plant for best economy, we assume that there 
are (n) units and then find what the total cost of production 
is in terms of (n). By plotting this equation, or by differen- 
tiating it and equating it to zero, we can determine the value 
of (n), the number of units required for minimum production 
cost, and thus the size of each unit. This is comparatively 
simple. 

But we can obtain better efficiency usually by using dif- 
ferent sizes of units rather than a number of units all of the 
same size. In such cases, fractional determination will give 
the desired results. That is, when the load curve consists of 
two or more distinct parts, one, let us say, of short duration 
but a heavy load, and the other of long duration and com- 
paratively light load, we can divide the curve into two parts, 
considering each separately. After we have determined thus 
the units best for each part of the load, we can then make such 
adjustments as will bring the two or more parts into harmony. 

145. Example 48. — A producer gas power plant carries a 
load in kw. as given by the equation 

M = 1000 — o.iiV. 

The cost per kw. of the units is given by the equation 

C = 50 - 0.005 u, 

where U is the size of the unit. 

The fuel cost per kw.h. at full load varies with the size of 
the unit thus : 

F = 0.007 — 0.0000006 U. 

At fractional loads, the per cent of full load efficiency ob- 
tained is 

P = 0.5(1 + X), 

where (X) is the per cent of full load. 

The attendance cost is constant at $0,003 P er kw.h. 



248 TYPE AND SIZE OF UNITS 

Assuming fixed charges at 10%, how many units of the 
same size will be required to give minimum production cost? 
Solution: (i) Total fixed charges per year are 
M cp = iooo X o.io X (50 — 0.005 U) 
= 5000 - 0.5 c/, 
where 

U = — ? , (n) being the number of units employed in the plant, 
n 

, TT 1000 
whence U = , 

n ' 



, ,, . 1000 s°° 

and M Q cp = 5000 — 0.5 -\ = 5000 — ^— 

dance cost per 

A = 0.003 / 1 



n 
(2) The total attendance cost per year is 

; MdN 
where 



/ 



M dN gives the total number of kw.h. produced per year 



/8760 
(1000 — o.iN)dN 

= 0.003 (1000N — o.o5A r2 ) 8760 
or A = 0.003 X 4,923,120 = $14,769.36. 

(3) Fuel Cost: If the unit runs at full load, the fuel cost is 
(F) per kw.h. given above. If it runs at x% of full load, the 



fuel cost is 



©-'■ 



(a) Case of one unit: 
In this case X = — - = 



M 1000 

since U = M = 1000 kw., 

so that P = 0.5(1 + o.ooiikf) 

and F = 0.007 — 0.0000006 X 1000 = 0.007 ~~ 0.0006, 

or F = 0.0064, 

and the actual fuel cost is 

, F 0.0064 0.1028 , , 

f = — = ; — r = ~ —r per kw.h. 

J P 0.5(1 + 0.001M) (1 + o.ooiikf)^ 



DETERMINATION OF NUMBER OF UNITS 249 

For the load (M) the fuel cost per hour is 

,,. 0.0128 M 

Mf = 



(1 + 0.001M) 

and the total annual fuel cost is 

.0128 M dN 



Jmjm -/* 



+ 0.001 M) 

But since M = 1000 — o.iiV, 

,, 1000 — M , ,,x 

N = = 10(1000 — M) 

0.1 

and dN = — 10 dM, 

MdM 



whence / Mf dN = — 0.128 / — 

J J 1000 V 1 



+ O.OO I if) 



so that 



fMfdM= , °' Y2 l \(i + o.ooiM)-log e (i + o.ooiM)? 24 

J (O.OOI) 2 L Jiooo 

= $38,400. 

Summary of Case (a), one unit. 
The fixed charges are 

5000 — - — = $4500. 

The attendance costs are $14,769.36, and the fuel cost is 
$38,400, whence the total annual production cost is 
4500 + 14,769.36 + 38,400 = $57,669.36. 

(b) Case of two units: 

In this case we have two units each of 500 kw. The first 
unit runs at fractional load from C to B, then at full load from 
B to E. The second unit cuts in at B, carrying the load area 
BDE. Or the units may carry the load area ABDO, both 
being equally loaded. 

Since M = 1000 — o.iiV, when M = 500, 
N = 5000 hours. 

The fuel cost for the load BC is 



/"8760 /•8760 17 

BC = / MfdN = M^-dN. 

1/5000 */5ooo -* 



250 



TYPE AND SIZE OF UNITS 



But when U = 500, 

F = 0.007 — 0.0000006 X 500.0 = 0.0067 
M 



and 



X = = 0.002 M, 

500 



whence BC = f ^f ; 

J 5000 0.5(1 + o 



= - 0.134 



002 M) 
MdM 



/*I24 

J 500 1 + 0.002 M 

or BC = - °' I3 \J (i + .oo2M)-\og e (i + .oo2M) 
\p.002y L 

= $9393.40. 



1 




So also 



BD= -0.134 / -—— 

^ool + O. 



Fig. 52. — Case of two units. 

dM 



whence 
and finally 



002 M 

= 7 °' T i 4 R 1 + 0.002M) - log c (i + o.oo2M)T 
(0.002) 2 l Js 

BD = $10,281.15 

/Sooo 
500 X 0.0067 dN 

]5ooo 
= $16,750. 



251 



UNITS IN A PLANT FOR GIVEN LOAD 

Summary of Case (b) : 
The total fuel cost is 

16,750 + 10,281.15 + 9393-40 = $36,424.55, 
and the total fixed charges are 

5000 -522 = S4750. 
2 

The total annual production cost is therefore 

36,424.55 + 475o + 14,769-36 = $55,943-91. 
(c) Case of Three Units: 

In the case of three units, each is of 333.33 kw. size. The 
first unit runs at fractional load until it reaches full load, after 
M 



,0 


F 




^\ 


: 


K 

.) 


X 





Fig. 53. — Case of three units. 



which it continues to run at full load, while the excess is taken 
care of by cutting in a second unit. When the load exceeds 
666.67, the third unit is cut in. In this case the fixed charges 



are 



5 ooo - ^22 = $4833.33. 



And as before, the attendance charges are $14,769.36. With 
units of 333.33 kw. size, the full load fuel cost per kw.h. is 
F = 0.007 — 0.0000006 X 333.33 
= 0.0068 
and P = 0.5(1 + X), 

M 



where 



333-33 



= 0.003M. 



252 TYPE AND SIZE OF UNITS 

So that P = 0.5(1 + 0.003 Af), 

, r _ F _ O.OO68 _ O.OI36 

P 0.5(1 + 0.003M) (1 + 0.003 M) 
For the load area EDHJ, the fuel cost is 

">-/«» -/e^fe*"- 

or £Z> = - 0.136 I -. — ■ — - 

J (1 + 0.003M) 

since If = 1000 — 0.1N. 

We obtain the limits of the above integration for the value 

of M, when N = 8760, or M = 124, whence 

£Z) = - 0.136 / 7 r— = - , ° NO (1 + .003M) 

6 A3.33 (1 + .003M) (.oo 3 ) 2 L V 5 

]I24 
= $3795-9i> 
333-33 

when M = 333.33* 

N = 6666.67 hours 
and M = 666.67, 

N = 3333-33 hour s- 

The fuel cost during the load area EJOI and GFIK is 
therefore respectively 

EI = 333-33 X 6666.67 X 0.0068 = $15,111.11 
and FG = 333-33 X 3333-33 X 0.0068 = $7555-55- 

The fuel cost during the load area FEK is 

FE = - O.I36 / 7 ; — — , 

^333.33(1 + 0.003*0/) 
so that 

FE = - 7^4-1(1 + 0.003M) - log (1 + 0.003 Jlf)T 

(0.003) 2 L J 333.33 

= $4637.90. 

And since the fuel cost during the load area CFG is equal to 
that during the FEK, we have 

CF = $4637.90. 



UNITS IN A PLANT FOR GIVEN LOAD 



253 



The total fuel cost is therefore 
3795.91 + 15,111.11 + 7555-55 + 2 X 4657-90 = S35, 738.37 
and the entire production cost is 

S35J38.37 + 14,769.36 + 4833-33 = $55>34i.o6. 

(d) Case of Four Units: 

In this event, each unit is of 250 kw. capacity. 

The fixed charges are 



;oo 



5000 - J — = S4875. 
4 

The attendance cost is as before 814,769.36. 




u K L 

Fig. 54. — Case of four units. 

To obtain the fuel cost, note that FD = IC = JB, when 
EF = 250 kw.. N = 7500 
when M = 750; N = 2500 

when M = 500; N = 5000. 

The fuel cost for the load areas FDGI plus GCHJ plus 
EBKO is 

250(2500 + 5000 + 7Soo)F, 



254 TYPE AND SIZE OF UNITS 

where F = 0.007 — 0.0000006 X 250 = 0.00685. 
Therefore this fuel cost is 

250 X 15,000 X 0.00685 = $25,687.50. 
So, also, the load areas EFD = DIC = CJB. The fuel 

cost during one of these load areas is, since X = — = 0.004M 

250 

f MfdN= f M x 0-006 85 dN C MdM 

J J J 0.5(1 + o. 



)685 dN 

- =-0.137 



r m_ 

.004M) ^ 0/ J (1 + 0.004M) 

= - ( -^^-2[( I + O.OO4M) ~ l0g e (l + .00 4 M)T 

= $2627.83. 

And for the three triangular areas, the fuel cost is 
3 X 2627.83 = $7883.49. 

Finally, for the load area BA LK, the fuel cost is 

BA = - °' 1 ^ 7 \{i + 0.004M) - log e (i + 0.004M)] 

= $1829.81. 

Whence the total fuel cost is 

25,687.50 + 7883.49 + 1829.81 = $35,400.80, 

and the total annual production cost is 

35,400.80 + 14,769-36 + 4875 = $57,045-16. 



~|I24 



TABLE 93 
General Summary 



Number of units 


Production cost 


I 

2 
3 

4 


$5 7, 669. 36 

55,943 -9* 
55,341.06 
57,045 16 



It is evident then that under the conditions assumed in the 
example, three units will give the best economy, while to have 



DESIGN OF PLANT FOR BEST ECONOMY 255 

four units — only one too many — gives nearly as poor economy 
as having only one. 

As compared with one unit, the three units save in total 
production cost 

57,669.36 - 55,341.06 = $2,328.30 

which capitalized at 5% equals the very neat sum of $46,566. 
Under the conditions assumed, then, a three-unit plant would 
be worth $46,566 more than the same plant with only one 
unit. And there would be very nearly this same difference in 
value between a four- and three-unit plant. But under the 
conditions assumed in this example, the entire plant under 
the worst conditions ( U = o) would have a first cost of only 
$50,000. Therefore, a proper choosing of the number of units 
has resulted in practically doubling the value of the plant! 

As a further illustration of the application of financial en- 
gineering to design of a system, we give the following com- 
paratively simple example. 

146. Example 49. — It is desired to install a pumping plant 
of 1000 g.p.m. capacity to pump against a static head of 50 feet. 
Length of pipe line required is 200 feet. Determine the sizes 
and types of each part of the system for the following operat- 
ing conditions: Length of service, 5 months at 10 hours per 
day, cost of electric power if used 2 cents per kw.h., cost of 
#2 distillate 6 cents per gallon, interest 5%. 

Solution: 1 (a) Belted Electric. — For this type of plant we 
will have: 

Cost of 6" belted centrifugal pump $150.00 

Cost of 25 h.p. 3 phase, 1800 r.p.m. motor with rails 

and pulley 390.00 

Cost of belt 2 1 .00 

Cost of building size 14' X 22' 345 -oo 

Cost of foundations 35-oo 

Total $941.00 

The power consumed based on a pump efficiency of 60% 
and belt efficiency of 85% will be 24.7 h.p. at the motor. 



256 TYPE AND SIZE OF UNITS 

i (b) Direct-connected Electric. — In this case, we will have: 

Cost of 6" centrifugal pump with sub-base and flexible 

coupling for direct connection $300.00 

20 h.p. 3 phase, 220 volt, 1200 r.p.m. motor less rails 

and pulley -. 410.00 

Foundations 20.00 

Cost of building n' X 14' 190.00 

Total $920.00 

The power consumed at the motor is guaranteed at 20 h.p. 

Comparison. — A comparison shows, at once, that the cost 
of this part of the installation is practically the same in either 
case, the reduced size of building fully offsetting in the latter 
case the increased cost of the direct connected pump. But 
the life of the direct connected pump and motor is consider- 
ably greater than the belted pump and motor, due to the 
absence of belt strains, while the life of the belt is always 
comparatively short. Besides this we have a saving in the 
latter case of approximately 5 h.p. costing ioe' per hour or 

0.10 X 10 X 150 days = $150 per year, 

having a vestance of $3000. 

There is therefore no question of our deciding in favor of 
the direct-connected over that of the belted outfit. In fact 
the latter would be rather a liability. 

(2) Oil Engine Plant. — In this case, we will have: 

Cost of 6" belted centrifugal pump $150.00 

" " 25 h.p. semi-Diesel oil engine. . . . 1200.00 

" " belt 22.00 

" " building 16' X 28' 500.00 

" " foundations 180.00 

Total cost $2052.00 

Cost of Operation — Engine Plant. — The full 25 h.p. will be 
required. At o.6# of fuel oil guaranteed per h.p.h., the 
engine will consume 15/ or 2 gallons of oil per hour costing 12 p. 



DESIGN OF PLANT FOR BEST ECONOMY 257 

This gives a cost of 0.12 X 10 X 150 days = $180 per year 
(5 mo.). 

Cost of attendance per year = $60 
" " oil, waste, and supplies = 50 

The total operating cost is then 

180 + 60 -f- 50 = $290. 

Cost of Operation, D.C. Electric Plant. — The output of the 
motor was 20 h.p. With a guaranteed efficiency of 89.5%, 
this gives an imput requirement of 22.34 h.p., costing 44.68 
cents per hour or 

0.4468 X 10 X 150 days = $670.20 per year (5 mo.) 
Cost of attendance = 33.00 
" " oil and waste = 21.60 
Total per year (5 mo.) = $724.80 

A comparison shows that the first cost of the engine plant 
is considerably over twice the first cost of the direct connected 
electric plant, but the cost of operation of the former is con- 
siderably less than half that of the latter. To determine the 
comparative value, and thus reach a decision, we proceed as 
follows : 

TABLE 94 
Direct Connected Electric Pumping Plant 



Item 


Cost (C) 


Life, years 


Term factor 


Deprec. 
vestance, 


Pump 


$300.00 

410.00 

20.00 

190.00 


25 

2 5 
Permanent 

15 


. 70469 
. 70469 
I . 00000 
0.51897 


$426 . 00 

581 .00 

20.00 


Motor 


Foundation 


Building 


367.00 
$1394.00 


Total depreciation 
vestance 











Cost of operation $724.80 per year, giving an operating vest- 
ance of 724.80 -f- .05 = $14,496. 

The total vestance is then 

$14,496 + 1394 = $15,890. 



258 



TYPE AND SIZE OF UNITS 



TABLE 95 



Oil Engine Plant. — 



Item 


Cost (C) 


Life, years 


Term factor 


Deprec. 
vestance 
C + T 


Pump 


$150.00 

1 200 . OO 

22.00 

500.00 

180.00 


15 

15 

5 

15 
Permanent 


O.51897 
O.51897 
O.25275 
0.51897 
1 . OOOOO 


$289.00 

2320.OO 

87.00 

966.00 


Engine 


Belt 


Building 

Foundation 


Total depreciation 
vestance 


$3842.00 











The total cost of operation per year was $290, giving an 
operating vestance of 

$290 + .05 = $5800. 

The total vestance is then 

3842 + 5800 = $9642 

as compared with a total vestance of $15,890 for the direct con- 
nected electric plant. The value of the engine plant as com- 
pared with the electric plant is as 15,890 to 9642 or 1.55. 
That is the engine plant is worth a little over 1.5 times as 
much as the electric plant. 

Should we use #1 distillate in the engine costing 8 cents per 
gallon, instead of the #2 distillate, our fuel cost per year would 
increase $60, corresponding to an increase in operating vest- 
ance of $1200. 

The total vestance under these conditions would be 

9642 -f 1200 = $10,842, 

remaining still far superior to the electric plant. Our deci- 
sion on this part of the plant must then be in favor of the oil 
engine pumping equipment. 

Discharge Piping. — The first problem here would be to 
decide on the class of pipe to be used, i.e., standard wrought 
iron, cast iron, riveted steel or wood pipe. We have been 



DESIGN OF PLANT FOR BEST ECONOMY 



259 



present many times when the representatives of the manu- 
facturers of each class have argued the merits of their own 
and sometimes the demerits of other classes. Such special 
pleading, however, gets us nowhere. The tendency of city 
councils is usually to buy as expensive pipe as they can or 
think they can afford. But the question is merely, which will 
render the service cheapest. 

In this case we have a very low head to pump against. 
Pipe strong enough to stand very high heads is therefore en- 
tirely unnecessary. The cost of operation of pipe is the cost 
of the » friction-horse-power consumed by the pipe. This is 
lowest for wood pipe. Under such conditions, we should de- 
cide on planed machine-banded wood pipe. 

We must now determine the proper size of this pipe as 
follows : 

We have found that the cost of operation per year (150 
days @ 10 hrs.) was $290 total. The cost per hour is then 

290 -^ 1500 = $0.1933, 
and the cost per h.p.h. is 

0.1933 * 25 = $0.00773, 
whence we determine as follows: 

TABLE 96 
Machine-banded Wood Pipe for iooo G.p.m. and 50" Head 



Size 
in. dia. 



6" 

8" 

10" 

12" 

14" 
16" 



Cost (C) 
per ioo' 


Life, 
years 


Depre- 
ciation 
vestance 


Friction 
head, ft. 
per 100' 


Friction 
h.p. 


Operating 
vestance 


$20.50 


25 


2Q.IO 


IO.3 


5 15 


$1198.00 


25-25 


25 


35-80 


2-51 


1-255 


292.00 


33 25 


25 


47.20 


O.83 


0.415 


96.20 


37-75 


25 


53-50 


0-34 


0. 170 


39 50 


48.00 


25 


68.10 


O.16 


0.080 


18.56 


62 .00 


25 


88.00 


O.O9 


0.045 


10.44 



Total 
vestance 
per ioo' 



$1227.10 

327.80 

143.40 

93.OO 

86.66 

98.44 



An inspection shows that the total vestance is a minimum for 
14" pipe, this size giving best financial efficiency. For 1000 
g.p.m. and prices, etc., as listed, this size is best, irrespective of 



260 



TYPE AND SIZE OF UNITS 



the length of the line. For more expensive pipe, a smaller 
size would give best financial efficiency. 

It has been customary in an installation of this type to use 
eight inch pipe as best. But it is evident that with this size of 
pipe, the conveyance of the water would cost us nearly four 
times as much as with fourteen inch pipe. 

The effect of increased power cost is to force us to use larger 
pipe, while decreased power cost permits us to use smaller 
pipe for best financial efficiency. It is just as great an error 
to use too expensive equipment as too cheap. 

We have now completed all of the plant except intake canal, 
screen and trash racks, suction pipe, check valve and hand 
primer, together with necessary increasers, bolts, gaskets, 
flanges, etc. 

These items, exclusive of the check valve, amount to $187.50. 

The check valve offers a special problem. If a six inch swing 
check is used, it will introduce a friction head of 5.1 feet cost- 
ing $12 per month for power. This amount is enormous, 
being over 10% of the total power required. The use of a 
larger check valve will help some as shown in the table below. 

TABLE 97 
Swing Check Valve, iooo G.p.m. Capacity 



Size 
in. dia. 


Cost (C) 


Life, 
years 


Depre- 
ciation 
vestance 


Friction 
head 


Friction 

h.p. 


Operating 
vestance 


Total 
vestance 


6" 

8" 

10" 

12" 

14" 


$18.90 
3I-50 
42.00 
63.00 
84.00 


25 
25 
25 
25 
25 


26.90 
44.90 
59.60 
89.50 
119.60 


5-1 

1-25 

O.41 
O. 17 
O.08 


2.65 

O.625 

O.255 

O.085 

O.04 


$615.00 

145.00 

59.20 

19.80 

9-30 


$641 . 90 
189.90 
118.80 
109.30 
128.90 



This analysis shows that the twelve inch check valve is best. 
But if this is so, then the discharge nozzle of a 1000 g.p.m. pump 
should be twelve inch instead of six inch, with a corresponding in- 
crease in the size of the diffuser and suction. The better thing to 
do would be to use a gate valve in which the friction is practically 



DESIGN OF PLANT FOR BEST ECONOMY 



261 



zero, when the gate valve is wide open as it will always be 
when the pump is in operation. The gate valve will serve 
perfectly well for priming as well as the check valve. In case 
of stoppage of the engine, there will be nothing to prevent a 
reversal of the flow in the pipe line. If the end of the dis- 
charge is above the water line in the reservoir or canal, this 
reversed flow can be limited to the water in the pipe line, and 
thus amount to very little. We should, in this installation, 
choose a six inch gate valve placed directly on the nozzle of the 
pump, costing $24.50, and having a total vestance of $34.40 
instead of $109.30 for the best check valve. 

TABLE 98 
Summary or Plant Items (All Costs Installed) 



Item 


Size 


First cost 


Depre- 
ciation 
vestance 


H.p. 


Operating 

cost per 

year 


Vestance 


Total 

vestance 


Pump 

Engine 

Belt 


6" 

25 h.p. 

8" 

16' X 28' 

12 yds. 

14" X 200' 

6" 


$150.00 

I 200 . OO 

22.00 

500 . OO 

180.00 
96.00 

24.50 
187.50 


$289.00 

2320.OO 

87.00 

966 . OO 

180.00 
136.20 

34-40 

211. 20 


21 


$243-60 


$4872.00 


$5161.00 

2320.OO 

957.00 

966 . 00 


3-75 


43-50 


870.00 


Building 

Foundation . . . 








Piping 

Gate valve. . . . 


0.08 


0-93 


18.60 


154.80 
34-40 


Incidentals. . . . 




















Totals 




$2360.00 


$4223.80 


$24.83 


$288.03 


$5750.60 


$9984.40 







There is still one item open to discussion in the above and 
that is the belt, this little item of $22 costing $957 in vestance. 
The only substitute we could here use would be a silent chain 
drive or reduction gear running in oil, together with a common 
sub-base for engine and pump. The efficiency of such transmis- 
sion is guaranteed at 98% and will cost $160 for the base and 
$120 for the transmission. Under these conditions we would 
have the following: 



262 



TYPE AND SIZE OF UNITS 

TABLE 99 



Item 


First cost 


Life, years 


Depre- 
ciation 
vestance 


Operating 

cost per 

year 


Operating 
vestance 


Total 

vestance 


Base 


$160.00 
120.00 


Permanent 
IS 


$160.00 
232.OO 






$160 OO 


Transmission .... 


$3-58 


$71.60 


303 . 60 


Total 


$280.00 




$39 2 -°° 


$3-58 


$71.60 


$463-60 







But with such an outfit we could reduce the size of the 
building from 16' X 28' to 14' X 18' costing $285, with a 
depreciation vestance of $440. This represents a saving of 
vestance on the smaller building required of 

966 — 440 = $526 
more than offsetting the vestance of the direct geared trans- 
mission and sub-base. We would therefore decide in favor of 
this drive. There is also a saving of two yards in foundation 
to be taken into account on this design. Under these condi- 
tions our plant items, costs, and vestances would be as follows: 

TABLE 100 
Final Summary of Plant 



Item 


Size 


First cost 


Depre- 
ciation 
vestance 


H.p. 

used 


Operating 

cost per 

year 


Operating 
vestance 


Total 

vestance 


Pump 

Engine 

Base 


6" 
25 h.p. 


$150.00 

1 200 . OO 

160.00 

1 20 . OO 

285.00 

150.00 

96.00 

24.50 

187.50 


$289.00 
2320.OO 
160.OO 
232.OO 
440 . OO 
150.OO 
136.20 
34-40 
211 .20 


21 


$243.60 


$4872.00 


$5161.00 
2320.OO 








Transmission. . 




0.05 


3.58 


71.60 


303.OO 
440 . OO 
150.OO 
154.80 
34-40 
211 .20 


Building 


14' X 18' 
ic yds. 

14" X 20o' 

6" 


Foundation . . . 








Piping 


0.08 


0.93 


18. 6c 


Incidentals . . . 




















Totals 




$2373-00 


$3972.80 


21.13 


$248.11 


$4962.20 


$8935.00 







We have by the substitution of the direct geared drive above 
reduced the total vestance by over $1000 and by so doing re- 
duced the cost of the service more than 10%. The above 
plant, the product of scientific cost analysis as applied to this 



DESIGN OF PLANT FOR BEST ECONOMY 263 

simple design, has a total vestance less than one-third as great 
as the average best plants installed, i.e., the total service cost 
of the plant is less than one-third of that found in the best 
plants of this size, and possibly one-fifth of the average plant 
installed. 

147. Example 50. — Design a 1100 kw. power plant assum- 
ing interest 5 %, land 50 cents per square foot, fuel oil $1.75 per 
barrel (320$), coal $4 per ton; on the basis of 18,500 B.t.u. 
per pound for the oil and 12,000 B.t.u. per pound of coal. 
The conditions of use are full load for 3000 hours per year. 

Solution: Eleven hundred kilowatts equals approximately 
1500 h.p. At 90% mechanical efficiency and direct drive, this 
would demand a 1700 i.h.p. engine. The question at once 
arises as to what type of an engine we should use. We have 
for consideration (1) internal combustion motor and (2) steam 
engine (a) condensing or (b) non-condensing and (3) steam 
turbine (a) condensing, (b). non-condensing. 

(1) Internal Combustion Motor. — Of this size we can con- 
sider with profit only the Diesel engine and the producer gas 
plant. 

(a) In the case of the Diesel plant, we have 

Cost of Diesel engine $120,000.00 

" " direct-connected generator 16,000.00 

" " piping 1,000.00 

" " building 6,000.00 

" " foundations 5,000.00 

Switchboard and auxiliaries 3,000.00 

Total $151,000.00 

The engines are guaranteed at 0.43$ per h.p.h. or two barrels 
per hour, costing $3.50. Whence the cost per year of the 
fuel is $10,500.00. 

Cost of attendance per year $2,000.00 

" " maintenance per year 1,350.00 

" " oil, waste, etc 1,650.00 

Total $15,500.00 

Operating cost per h.p. year, $10.33!. 



264 



TYPE AND SIZE OF UNITS 



Assuming the life of the Diesel engine, switchboard, and 
auxiliaries at 15 years, that of the generator, building, and 
piping at 25 years, we may tabulate as follows: 



TABLE 101 
Summary — Diesel Engine Plant 



Item 


First cost 
erected 


Depreciation 
vestance 


H.p. 


Operating 

cost per 

year 


Operating 
vestance 


Total 
vestance 


Engines. . . . 
Generator. . 

Piping 

Building 

Foundation. 


$1 20,000.00 
16,000.00 
1,000.00 
6,000.00 
5,000.00 
3,000.00 


$231,230.00 
22,700.00 
1,420.00 
8,520.00 
5,000.00 
5,790.00 








$231,230.00 
332,700.00 
1,420.00 
8,520.00 
5,000.00 
9,923.00 


1500 


$15,500.00 


$310,000.00 














Auxiliaries. . 


20 


206.67 


4,133-00 


Totals. . . 


$151,000.00 


$274,660.00 


1520 


$15,706.67 


$314,133.00 


$588,793-00 



(b) For the producer gas plant, we would have: 

Cost of engines erected $75,000.00 

producers " 10,800.00 

generator " 16,000.00 

piping, starter, switchboard, etc 7,000.00 

building 10,500.00 

foundations 6,500.00 

Total $125,800.00 

The producer plant is guaranteed to produce a br.h.p.h. at 
the engines on 0.75$ of coal or a total of 11 25$ per hour, or 
5.625 tons per day of 10 hours. Allowing a standby loss of 
4%, this would mean a total consumption of 5.85 tons per day 
or 1755 tons per year, costing $7022 per year. 

For operating costs per year, we have: 

Fuel cost $7,022.00 

Attendance 10,000.00 

Maintenance 1,800.00 

Oil, waste, etc 2,078.00 

Total $20,900.00 

Cost per h.p. year = Si 



o-93i 



DESIGN OF PLANT FOR BEST ECONOMY 265 

Assuming a life of 20 years on the engine and producer, 
25 years on the generator and building and 15 years on the 
starter, switchboard, etc., we may summarize as follows: 



TABLE 102 
Summary Producer Gas Plant 



Item 


First cost 


Depreciation 
vestance 


H.p. 


Operating 

cost per 

year 


Vestance 


Total 
vestance 


Engine and 

producer . 

Generator. . 

Auxiliaries . . 

Building 

Foundation . 


$85,800.00 
16,000.00 

7,000.00 
10,500.00 

6,500.00 


$137,500.00 

22,700.00 

13,500.00 

14,900.00 

6,500.00 






$137,500.00 

440,700.00 

19,073.00 

14,900.00 

6,500.00 


1500 
20 


$20,900.00 
278.67 


$418,000.00 
5,573-oo 










1520 






Totals. . . 


$125,800.00 


$195,100.00 


$21,178.67 


$423,573.00 


$618,673.40 



A comparison shows that under the conditions assumed, 
the comparative value (valuance) of the Diesel Plant is 5% 
greater. We should therefore decide in favor of the latter by 
a very small margin. 

(c) We have now to consider the steam plant. For this 
type of plant, we have costs as follows: 



Cost of boilers $13,500 

" turbo-generators 20,500 

" building 10,500 

" foundations . 6,500 

" piping 6,000 

" stokers 7,500 

" ash conveyor 1,000 

" coal " 7,500 

" condensers 5,600 

" auxiliaries 2,500 

" stock and breeching 3,400 

Total $84,500 



Life in Years 

25 

25 

25 

Permanent 

25 
i5 
15 
i5 
25 
15 
15 



266 TYPE AND SIZE OF UNITS 

If the plant is guaranteed to produce a h.p.h. on 1.5$ of 
this coal, then the consumption per hour will be 2 25o#, to 
which must be added 10% for standby losses and auxiliary 
machinery, making a total of 247 5# or 1.2375 ton per hour. 
This means a consumption of 3700 tons per year at $4 per ton. 

We have then: 

Cost of fuel per year $14,800 

" " attendance 10,000 

" " maintenance 1,200 

" " oil, waste, etc 1,500 

Total $27,500 

Cost per h.p. year = $18. 33J. 

We may then summarize as follows: 

TABLE 103 

Summary of Steam Plant 



Item 


First cost 


Boilers 


$13,500.00 


Turbo-gener- 




ators 


20,500.00 


Building 


10,500.00 


Foundations. 


6,500.00 


Piping 


6,000.00 


Stokers 


7,500.00 


Ash convey- 




ors 


1,000.00 


Coal convey- 




ors 


7,500.00 


Condensers. . 


5,600.00 


Auxiliaries. . . 


2,500.00 


Stack, etc. . . 


3,400.00 


Totals 


$84,500.00 



Depreciation 
vestance 



$19,200.00 

29,200.00 

14,900.00 

6,500.00 

8,520.00 

14,500.00 

1,930.00 

14,500.00 
7,960.00 
4,820.00 
6,570.00 



$128,600.00 



H.p. 



1500 



90 



I590 



Operating 

cost per 

year 



$27,500.00 



1,650.00 



$29,150.00 



Vestance 



$550,000.00 



33,000.00 



$583,000.00 



Total 
vestance 



$19,200.00 

579,200.00 

14,900.00 

6,500.00 



85,230.00 



6,570.00 



$711,600.00 



Comparison of this plant with the Diesel shows that 
(a) The Diesel plant has a first cost 1.79 times as much 
(6) But its comparative value is 21 % greater than the steam 
plant. 



PROBLEMS 267 



PROBLEMS 

1. Design a steam plant of 1000 h.p. size, to operate at full load for 3000 
hours per year, so that each part will operate at maximum financial 
efficiency, assuming interest rate 5%, coal $4 per ton, with a heat value 
of 5.4 h.p.h. per pound. 

2. A pipe is to carry io,ooo# steam per hour to a condensing steam engine. 
The heat costs 0.15 cent per h.p.h. Taking into consideration the loss in 
radiation and the loss of available heat due to drop in pressure, determine 
the proper size of pipe for maximum financial efficiency. 

3. Taking the costs of pipe as given, and taking into consideration 
radiation and pressure losses, determine the pressure of steam to be used 
for conveying heat to a condensing engine for best financial efficiency 
irrespective of other than transmission considerations. 

4. A small electric generating plant is to carry a 50 kw. load for 2000 
hours per year and a 25 kw. load for another 1000 hours per year. Design 
the plant for best financial efficiency, assuming interest rate 5%, coal 
(5.4 h.p.h. per lb.) at $5 per ton, fuel oil (7.3 h.p.h. per lb.) at $1.85 per 
barrel, distillate (7.7 h.p.h. per lb.) at 25 cents per gallon. 

5. In problem (4), what would be the average cost per kw.h. produced? 
What would be the true cost per kw.h. during each period? 

6. A steam turbo-generating plant carries a load as given by the 
equation 

M = 1000 + o.$N. 

Taking into consideration the variation in efficiency of turbo-generators 
with size and load, and all other costs, determine the number of units 
required for maximum financial efficiency. 

7. In problem 6, if the plant consisted of one unit, how would it compare 
in value with best design? 

8. It is desired to install a pumping unit to deliver 5000 g.p.m. during 
May, June, and July and 2000 g.p.m. during August and September, against 
a fixed discharge head of 50 feet. Centrifugal pumps electrically driven are 
to be used. The pumps are set to stand 10 feet above the water at the be- 
ginning of the pumping season. From then to August 1st the suction heat 
increases uniformly to 20 feet and then remains constant until the end of 
the season. The power costs 1.5 cents per kw.h. Choose the type, size 
and number of units for best financial efficiency. 



INDEX 



Account, stock, 23 

Annual operating costs, 53-54 

Attendance, 7, 15, 22 

— cost of, centrifugal pumps, 91 
gas engines, 105 

generators, 99 

motors, 99 

oil engines, 105 

steam engines, 75 

Basic costs, 59-62 
Basis of rates, 3-4, 187 
Belts, vestance of, 174, 175 
Boilers, 75-76 

— price of, 77-78 
Buildings, cost of, 78-80 

Canals, cost of, 113-114 
Capitalization of earnings, 11,12 
Centrifugal pumps, attendance cost of, 

— efficiency of, 89-91 

— price of, 81-88 

— vestance of, 162-170 

Change point, 2, 140, 141, 145, 188, 189 

Charges, fixed, 6, 7 

Charges, replacement, 20 

Collection, 6 

Comparative value, 11, 29-30 

Comparison of power units, 158-162 

Competition, 2, 10 

Complete steam installations, cost of, 80 

Compound interest, 25-26 

Cost, 2, 4, 6, 14 

— analysis, 1-2 

— basic, 59-62 

— determination, 45 

— operating, 6, 7 

— original, n, 12 

— price (See price) 



— segregation, 1 

— of buildings, 78-80 

— of canals, 11 3-1 14 

— of complete steam installations, 80 

— of dams, 112 

— of hydro-electric installations, 115- 
116 

— of tunnels, 113 

— of replacement, 11, 12 

— of excavations, 114 

Dams, cost of, 112 
Depreciation, 7, 15, 16, 42 

— rates, 36-40 

— reserve, 18 

— vestance, 46-49 

Design for best economy, 5, 247-271 
Diesel engines, attendance cost of, 105 

— efficiency of, 106 

— price of, 101-102 

— vestance of, 150-152 

Direct current motors (See motors) 
Distribution, 6, 23 

Earnings, capitalization of. 11, 12 
Efficiency of centrifugal pumps, 89-91 

— Diesel engines, 106 

— gas engines, 104-105 

— generators, 98-99 

— motors, 96-97, 99 

— oil engines, 104-105 

— steam engines, 72-74 
Engines (See gas, steam, Diesel) 
Equation of load curve, 226-228 
Equipment, life of, 19, 20 
Equity, 27-30 
Excavations, 114 

Factor, term, 31, 35 

Financial efficiency, maximum, 224, 239 



269 



270 



INDEX 



Fixed charges, 6, 7 

Fractional loads, operation at, 224, 246 

Friction of pipe, no, 112 

Gas engines, attendance cost of, 105 

— efficiency of, 104-105 

— price of, 101-103 
Generators, attendance cost of, 99 

— efficiency of, 98-99 

— price of, 95-96 

— vestance of, 156-158 



— efficiency of, 104-105 

— price of, 100 

— vestance of, 146-150 
Operating costs, 6, 7, 15 

— annual, 53-54 

— life, 21 

— of equipment (See efficiency, attend- 
ance) 

— variable, 222-223 
— vestance, 35 

Original cost, n, 12 



Hydro-electric installations, n 5-1 16 
Heat transmission, 232 

Inadequacy, 17, 18 
Individual systems, 8 
Induction motors (See motors) 
Installations, complete steam, cost of, 80 
Installations, hydro-electric, n 5-1 16 
Instantaneous unit cost, 215-216 
Insurance, 7, 15, 21, 54-56 
Integral systems, 8 
Interest, 7, 15-16, 25 

— compound, 25-26 

Law of supply and demand, 10, n 
Life, natural, 21 

— operating, 21 

— of structures and equipment, 19, 20 
Load curve, equation of, 226-228 

— replot, 217, 218 

Maintenance, 7, 15, 22 

Market value of outstanding liabilities, 

11, 12 
Materials consumed, 7, 15, 21, 22 
Maximum efficiency, design for, 247-271 
Minimum costs, 224 
Motors, attendance cost of, 99 

— efficiency of, 96, 97, 99 

— price of. 92-95 

— vestance of, 153-156, 186 

Natural life, 21 

Number of units in a system, 243-247 

Obsolescence, 17 

Oil engines, attendance cost of, 105 



Pipe, price of standard, 107 

casing, 108 

riveted steel, in 

wood, 109, in 

— friction of, no, 112 

— vestance of standard, 1 76-181 
wood, 182-183 

Power, 7, 15 

— units, comparison of, 158-162 
Present worth of a depreciating equip- 
ment, 43-44 

Price, 2, 4, 13-14 
Price of boilers, 77-78 

— centrifugal pumps, 81-88 
— Diesel engines, 101-102 

— gas engines, 101-103 

— generators, 95-96 

— motors, 92-95 

— oil engines, 100 

— producer gas engines, 101-103 

— steam engines, 62-63 

— transformers, 96 

— turbo-generators, 64 
Prices, table of, 1 18-123 
Principal, 26 

Profits, 2, 13-14 

Production, 6 

Producer gas engines, prices of, 101-103 

Public utilities, 4, 7-10 

Pumps (See centrifugal) 

Rate, 2, 4 

Rates, basis of, 2-4 

— depreciation, 36-40 

— of electric power for Oregon, 184-185 

— of interest, 16 



INDEX 



271 



Records, 60 

Rents, 7, 15 

Repair, 7, 15 

Replacement costs, 11, 12, 20 

Replot, 217, 218 

Reserve, depreciation, 18 

Reserve units, 242 

Sale, 23 

Segregation, cost, 1 

Service modulus, 213-214 

Sinking fund, 18 

Size of system, 232-240 

Size of units, 244-247 

Stand-by units, 242-243 

Steam engines, attendance cost of, 75 

— complete installations, cost of, 80 

— efficiency of, 72-74 

— price of, 62-63 

— steam consumption, 65-74 

— vestance of, 134-144 
Stock account, 23 
Storage, 33 

Structures, life of, 19-20 
Supply and demand, law of, 10-11 
Systems, 8 

Table of depreciation rates, 40 

— prices, 1 18-123 

— term factors, 32 
Taxes, 7, i$, 21, 54, 55 
Term factor, 31-35 

Thermal efficiency (See efficiency) 
Time element in vestance, 127, 128 
Total vestance, 50-52 
Transformers, price of, 96 



Transmission, 6 
Transportation, 6 
Tunnels, cost of, 11 3-1 14 
Turbo-generators, price of, 64 

Unit cost, one period, constant load, 

187-190 
variable load, 200 

— two periods, constant load, 200-209 
variable load, 209-210 

— three periods — , 210-212 
, 212-213 

— continuously varying load, 215-216 
Units, stand-by, 242-243 

— number of, 243-247 
Uselessness, 18 
Utilities, public, 9, 10 

Valuance, 128 
Value, 3 

— comparative, 29-30 
Variable operating costs, 222-223 
Vestance, 45, 126, 129-134 

— depreciating, 46-49 

— of belts, 174-175 

— of centrifugal pumps, 162-170 

— of Diesel engines, 150-152 

— of generators, 156-158 

— of motors, 153-156, 186 

— of oil engines, 146-150 

— of standard pipe, 1 76-181 

— of steam engines, 134-144 

— of wood pipe, 182-183 

— operating, 35, 49 

— time element in, 127-128 

— total, 50-52 



